SCI 101 — General Physics I
Complete Study Guide — Data Science Pre-Master Program
Built from the course lecture slides (Weeks 1–5, 7–11) · Color-coded for exam revision
How to use the colors
🔵 Blue = Definitions & core concepts | 🟢 Green = Laws, rules & steps |
🟡 Yellow = Very important / attention | 🔴 Red = Warnings, exceptions, common mistakes |
🟣 Purple = Key terms & vocabulary
This chapter introduces the basic quantities used to describe motion along a straight line: position, velocity, speed, and acceleration — and how they relate to each other for constant acceleration and free fall.
1.1 Position, Velocity and Speed of a Particle
Definition
Speed is a scalar quantity — it does not include direction. It is the distance covered divided by the time it takes.
Definition — Average Velocity
Average velocity is the total displacement (change in position) divided by the elapsed time.
Average velocity v̄ = (x₂ − x₁) / t = Δx / Δt
Example 2-1 — Runner's average velocity
During a 3.00 s interval, a runner's position changes from x₁ = 50.0 m to x₂ = 30.5 m.
v̄ = (x₂ − x₁)/t = (30.5 − 50.0)/3 = −6.50 m/s
1.2 Instantaneous Velocity and Speed
Definition
Instantaneous velocity is the average velocity in the limit as the time interval becomes infinitesimally short (i.e., as Δt → 0).
Instantaneous velocity v = limΔt→0 (Δx/Δt) = dx/dt
Important
The instantaneous speed always equals the magnitude of the instantaneous velocity — but it equals the average velocity only if the velocity is constant.
Key terms
Constant velocity: moving in a straight line at the same speed. Changing velocity: braking, speeding up, or turning a corner. A change in speed or direction is called acceleration.
Acceleration
Definition
Acceleration is the rate of change of velocity.
Average acceleration a = (v₂ − v₁)/t = Δv/Δt
Instantaneous acceleration a = limΔt→0 (Δv/Δt) = dv/dt
Example — Average acceleration
A car accelerates along a straight road from rest to 25 m/s in 5.0 s. Find the average acceleration.
a = (v₂ − v₁)/t = (25 − 0)/5 = 5 m/s²
Example — Accelerating to 60 km/h
What is the acceleration if a car accelerates from stop to 60 km/h in 5 seconds?
a = (v₂ − v₁)/t = (60 − 0)/5 = 12 m/s²
Common confusion
If velocity is zero, acceleration is not necessarily zero — and if acceleration is zero, velocity is not necessarily zero (think of a ball at the top of its throw vs. an object moving at constant velocity).
1.3 Particles Under Constant Acceleration
Example — Car slowing down
A car moving right along the +x axis brakes. Initial velocity v₁ = 15.0 m/s, and it takes 5.0 s to slow to v₂ = 5.0 m/s.
a = (v₂ − v₁)/t = (5 − 15)/5 = −2 m/s²
Don't confuse
Negative acceleration means acceleration in the negative direction of the chosen coordinate system.
Deceleration means the acceleration is opposite in direction to the velocity (the object is slowing down). They are not the same thing — negative acceleration with a leftward-moving object actually speeds it up.
Law / Relationship
On a velocity–time (v vs. t) graph: the average acceleration over Δt = t₂ − t₁ is the slope of the straight line joining the two points (aav = Δv/Δt). The instantaneous acceleration at time t₁ is the slope of the v–t curve at that instant.
Example — Position given as a function of time
A particle's position is x = (2.10 m/s²)t² + (2.80 m).
- Average acceleration from t₁ = 3.00 s to t₂ = 5.00 s:
v(t) = dx/dt = 4.20t → v₁ = v(3) = 12.6 m/s, v₂ = v(5) = 21 m/s
a = (v₂ − v₁)/(t₂ − t₁) = (21 − 12.6)/(5 − 3) = 4.20 m/s²
- Instantaneous acceleration as a function of time:
a = dv/dt = d/dt(4.20t) = 4.20 m/s² (constant)
Building the constant-acceleration equations
Average velocity over the interval: v̄ = Δx/Δt = (x − x₀)/(t − t₀).
Acceleration (assumed constant): a = (v − v₀)/(t − t₀).
Because velocity increases at a constant rate, the average velocity is also the mean of initial and final velocity:
Mean velocity under constant acceleration v̄ = (v₀ + v)/2
Combining the equations
Combining the relationships above gives the full set of constant-acceleration ("kinematics") equations used throughout this course:
Position vs. time x = x₀ + v₀t + ½at²
Velocity vs. time v = v₀ + at
Velocity vs. position (eliminates t) v² = v₀² + 2a(x − x₀)
1.4 Freely Falling Objects
Definition
Near the Earth's surface, all objects experience approximately the same acceleration due to gravity, g ≈ 9.80 m/s², directed downward — regardless of mass, as long as air resistance is ignored.
Example — Falling from a tower
A ball is dropped (v₀ = 0) from a tower 70.0 m high. How far has it fallen after t₁ = 1.00 s, t₂ = 2.00 s, t₃ = 3.00 s?
x = x₀ + v₀t + ½at² = ½(9.8)t² = 4.9t²
- x(1) = 4.9(1)² = 4.9 m
- x(2) = 4.9(2)² = 19.6 m
- x(3) = 4.9(3)² = 44.1 m
Example — Ball thrown upward at 15 m/s
(a) Time to reach maximum height (velocity = 0 at the top):
v = v₀ + at → 0 = 15 + (−9.8)t → t = 15/9.8 = 1.53 s
(b) Maximum height:
y = y₀ + v₀t + ½at² = 0 + 15(1.53) + ½(−9.8)(1.53)² = 11.48 m
(c) Using v² = v₀² + 2a(y_f − y_i), with initial and final position both zero (ball returns to the launch point):
v² = (15)² − 2(9.8)(0−0) = 225 → v = ±15 m/s
🎯 Exam Focus — Chapter 1
- Know the difference between speed (scalar) and velocity (has direction), and between negative acceleration and deceleration — this is a classic trick question.
- Memorize all three kinematics equations and know which one to use depending on which variable is missing (no time → use v² = v₀²+2a(x−x₀)).
- In free-fall problems, always define a sign convention first (e.g., up = positive) and keep it consistent for v₀, v, y, and g.
- Velocity = 0 at the very top of a throw — but acceleration is still g the whole time (never zero) during free fall.
- On a v–t graph, slope = acceleration; on an x–t graph, slope = velocity.
This chapter covers the difference between vectors and scalars, how to add/subtract vectors, how to break them into components, and the two ways to multiply vectors: the dot product and the cross product.
2.1 Vectors and Scalar Quantities
Definition
A vector has both magnitude and direction. A scalar has only magnitude.
| Vector quantities | Scalar quantities |
|---|
| Displacement, Velocity, Force, Momentum | Mass, Time, Temperature |
2.2 Basic Vector Arithmetic
Graphical addition
In one dimension, simple addition and subtraction (with correct signs) are all that's needed. In two dimensions, if the paths are at right angles, the resultant displacement can be found using the Pythagorean theorem. Adding vectors in the opposite order gives the same result (vector addition is commutative).
Negative of a vector
The negative of a vector is the vector that, when added to the original, gives a resultant of zero. It has the same magnitude but points in the opposite direction. This is how vector subtraction is defined.
2.3 Components of Vectors and Unit Vectors
Key term — Unit vector
Unit vectors (î, ĵ, k̂) form a set of mutually perpendicular vectors in a right-handed coordinate system, each with magnitude 1: |î| = |ĵ| = |k̂| = 1.
Any vector can be expressed as the sum of two (or three) perpendicular components. If the components are perpendicular, trigonometric functions can be used to find them.
Components from magnitude and angle aₓ = a cos θ and aᵧ = a sin θ
Magnitude and angle from components a = √(aₓ² + aᵧ²) and tan θ = aᵧ/aₓ
where θ is the angle the vector makes with the positive x-axis, and a is the vector's length. Components fully define a vector. Angles may be measured in degrees or radians (a full circle = 360° = 2π rad).
2.4 Multiplying Vectors
Vector × scalar
Multiplying a vector z by a scalar c gives a new vector with magnitude |c| times the original, pointing the same direction as z (or opposite, if c is negative). To multiply by a scalar, multiply each component by c; to divide, multiply by 1/c.
Example
Multiply vector z = −3î + 5ĵ by 5:
5z = −15î + 25ĵ
The dot (scalar) product
Results in a scalar. a and b are magnitudes, φ is the angle between the two vectors' directions. It is commutative: a·b = b·a.
Dot product (magnitude form) a⃗·b⃗ = ab cos φ
Dot product (component form) a⃗·b⃗ = (aₓî+aᵧĵ+a_zk̂)·(bₓî+bᵧĵ+b_zk̂) = aₓbₓ + aᵧbᵧ + a_zb_z
Interpretation
A dot product equals the magnitude of one vector times the scalar component of the other vector in the direction of the first (i.e., the projection of one onto the other, times the magnitude).
The cross (vector) product
Results in a new vector, with a direction perpendicular to both original vectors, found using the right-hand rule (place vectors tail-to-tail, sweep fingers from the first to the second, thumb points in the direction of the resultant).
Cross product magnitude c = ab sin φ
Cross product component expansion a⃗×b⃗ = (aᵧb_z − bᵧa_z)î + (a_zbₓ − b_zaₓ)ĵ + (aₓbᵧ − bₓaᵧ)k̂
Common mistake
The cross product is NOT commutative: b⃗×a⃗ = −(a⃗×b⃗). Also, if a⃗ and b⃗ are parallel or antiparallel, a⃗×b⃗ = 0. The magnitude |a⃗×b⃗| is maximum when a⃗ and b⃗ are perpendicular.
Example — Cross product
Find A×B where A = (1,0,1) and B = (−1,0,1):
A×B = (0·1 − 0·0)î − (1·1 − 1·(−1))ĵ + (1·0 − 0·(−1))k̂ = 0î − 2ĵ + 0k̂
A×B = (0, −2, 0)
🎯 Exam Focus — Chapter 2
- Dot product → scalar answer. Cross product → vector answer. Don't mix them up.
- Dot product uses cos φ; cross product uses sin φ — a very common mix-up.
- Cross product is anti-commutative (order matters); dot product is commutative (order doesn't matter).
- Practice converting between (magnitude, angle) form and (component) form in both directions.
- Right-hand rule direction for cross products is a frequent source of sign errors — practice it physically with your hand.
Vectors are now used to describe motion in a plane. Two important special cases are studied in depth: projectile motion and uniform circular motion.
3.1 Position, Velocity and Acceleration (Vector Form)
Definition
Position is measured from the origin of a coordinate system. The displacement vector is the difference between two position vectors — it does not depend on the origin, unlike position itself.
Position & velocity for constant acceleration (vector form) r⃗ = r⃗₀ + v⃗₀t + ½a⃗t² | v⃗ = v⃗₀ + a⃗t
Average velocity = displacement ÷ time interval. Instantaneous velocity is the limit of average velocity as the time interval shrinks to zero — its direction is always tangent to the particle's path. Average acceleration = change in velocity ÷ time interval; instantaneous acceleration is the corresponding limit.
Example — 2D motion with constant acceleration
A particle starts from the origin at t = 0 with initial velocity components v
xi=20 m/s, v
yi=−15 m/s, and acceleration a
x=4 m/s², a
y=0.
(A) Total velocity vector at any time t:
vx = vx0 + axt = 20 + 4t
vy = vy0 + ayt = −15 + 0 = −15
v⃗(t) = (20 + 4t)î − 15ĵ
(B) Velocity at t = 5.0 s and its angle with the x-axis:
v⃗(5) = (20+4×5)î − 15ĵ = 40î − 15ĵ
tan θ = v_y/v_x = −15/40 → θ = tan⁻¹(−15/40) = −21°
(C) Speed of the particle (magnitude of v⃗):
speed = |v⃗| = √(40² + (−15)²) = 43 m/s
3.2 Projectile Motion
Definition
A projectile is an object moving under the acceleration of gravity alone; its path lies in a vertical plane. It is launched with initial speed v₀ at angle θ₀.
Key idea
Horizontal and vertical motions are independent: the x-component of velocity is constant (no horizontal acceleration); the y-component follows constant-acceleration equations with a = −g.
Horizontal motion (constant velocity) x − x₀ = v₀ₓt = (v₀ cos θ₀)t
Vertical motion (acceleration = −g)
y − y₀ = v₀ᵧt − ½gt² = (v₀ sin θ₀)t − ½gt²
vy = v₀ sin θ₀ − gt
vy² = (v₀ sin θ₀)² − 2g(y − y₀)
Horizontal range R = (v₀²/g) sin 2θ₀
Important
The horizontal range R is maximum when the launch angle is 45°.
Example — Long jumper
A long jumper leaves the ground at speed 11.00 m/s at some launch angle above the horizontal.
(A) Horizontal distance and (B) maximum height are found directly from the range formula and the vertical-motion equations above, using the given launch angle and speed.
Example — Stone thrown from a building (Figure 4.14)
A stone is thrown from the top of a 45.0 m building at 30.0° above the horizontal, with initial speed 20.0 m/s.
(A) Time to reach the ground: solving the vertical-motion quadratic in t gives t = 4.22 s (ignoring the negative root).
(B) The speed just before striking the ground is then found from vy = v₀ sin θ₀ − gt combined with the constant vx, and combining components to get the resultant speed.
3.3 Analysis Model: Particle in Uniform Circular Motion
Definition
A particle is in uniform circular motion if it travels around a circle (or circular arc) at constant speed. Because the direction of velocity keeps changing, the particle is still accelerating — velocity and acceleration have constant magnitude but changing direction. The velocity vector is always tangent to the path; the acceleration vector always points toward the center.
Key term
This acceleration is called centripetal acceleration ("center-seeking") — it is directed radially inward.
Centripetal acceleration a = v²/r
Period of revolution T = 2πr/v
Angular speed ω = 2π/T
Example — Angular velocity
A particle in a circular orbit rotates at 200 times per minute. Find its angular velocity.
T = 60/200 = 0.3 s
ω = 2π/T = 2π/0.3 = 20.93 rad/s
🎯 Exam Focus — Chapter 3
- For projectile motion, always split the problem into independent x and y parts — this is the single most important idea in the chapter.
- Remember vx is constant throughout the flight (no horizontal force); only vy changes, due to gravity.
- Range formula R = (v₀²/g)sin2θ₀ only applies when launch and landing heights are equal — if not (like the stone off the building), you must solve the full vertical-motion equation for t.
- In uniform circular motion, speed is constant but velocity is NOT (direction always changes) — this is why the object still accelerates.
- Don't confuse tangential velocity (v) with angular velocity (ω); they are related by v = ωr (see Chapter 7 for the full linear–rotational correspondence).
This chapter explains why motion changes: the role of force and mass, Newton's three laws, weight, normal force, tension, and friction.
4.1 Concept of Force
Definition
A force is a push or pull acting on an object, and it causes acceleration. Dynamics studies the causes of motion, addressed through the forces acting on an object and its mass.
Characteristics of forces
Unit: the newton (N), where 1 N = 1 kg·m/s². Acceleration is proportional to the net (applied) force. Forces are vectors. Net force is the vector sum of all forces on an object (principle of superposition).
Limits of Newtonian mechanics
Newtonian mechanics is valid for everyday situations, but not valid for speeds an appreciable fraction of the speed of light, or for objects at the atomic scale.
4.2 Newton's First Law and Inertial Frames
Newton's First Law
In the absence of external forces, and viewed from an inertial reference frame, an object at rest remains at rest, and an object in motion continues in motion with constant velocity (constant speed in a straight line). Newton's first law is not true in all frames — an inertial frame is one in which Newton's laws hold.
4.3 Newton's Second Law
Newton's Second Law
The net force on a body equals the product of the body's mass and acceleration.
Newton's second law F⃗net = ma⃗
Component form (axes are independent) Fnet,x = max, Fnet,y = may, Fnet,z = maz
Mass
Mass is the characteristic of a body that relates a force on it to the resulting acceleration — a measure of resistance to a change in motion. Mass is not the same as weight, density, or size. Acceleration is inversely proportional to mass.
Example — Same force, different masses
An 8.0 N force is applied to bodies of different masses:
| Mass | Resulting acceleration |
|---|
| 1 kg | 8 m/s² |
| 2 kg | 4 m/s² |
| 0.5 kg | 16 m/s² |
System
A system consists of one or more bodies. A force on bodies inside a system, exerted by bodies outside it, is an external force. Net force on a system = sum of external forces. Forces between bodies within the system are internal forces — they are not included in a free-body diagram (FBD) of the system, since internal forces cannot accelerate the system as a whole.
Example — Hockey puck struck by two forces
A 0.30 kg puck on frictionless ice is struck by two forces: F₁ = 5.0 N at θ = 20° below the x-axis, and F₂ = 8.0 N at φ = 60° above the x-axis.
ΣFx = F₁cosθ + F₂cosφ ΣFy = F₁sinθ + F₂sinφ
ax = ΣFx/m ay = ΣFy/m
Substituting numbers:
ax = [(5.0)cos(−20°) + (8.0)cos(60°)]/0.30 = 29 m/s²
ay = [(5.0)sin(−20°) + (8.0)sin(60°)]/0.30 = 17 m/s²
Magnitude and direction of the acceleration:
a = √(29² + 17²) = 34 m/s²
θ = tan⁻¹(ay/ax) = tan⁻¹(17/29) = 31°
4.4 Gravitational Force and Weight
Gravitational force
A pull that acts on a body, directed toward a second body (usually the Earth). In free fall (no air drag):
Gravitational force Fg = mg (vector form: F⃗g = −Fgĵ = m g⃗)
Weight
The magnitude of the gravitational force acting on an object is its weight, W = Fg (with the ground as the inertial reference frame).
Mass–weight relationship W = mg
Example — Weight of a 70.0 kg person
W = mg = 70.0 × 9.8 = 686 N
Mass vs. Weight
Mass is the amount of matter in an object — it does not change with location. Weight is caused by gravity and depends on location: the farther from a planet, the smaller the gravitational force; a more massive planet produces more weight for the same mass.
4.5 Normal Force, Friction, and Tension (Forces You'll See in Diagrams)
Normal force
When a body presses against a surface, the surface pushes back with a normal force (perpendicular to the surface) — this is what happens when you stand on a surface.
Friction (introductory)
Occurs when one object slides, or attempts to slide, over another; directed along the surface, opposite to the direction of intended motion.
Tension
A cord pulled taut exerts a force T on the body it's attached to, directed along the cord. An ideal (massless, unstretchable) cord pulls on both ends with the same tension T — even around a massless, frictionless pulley.
4.6 Newton's Third Law
Newton's Third Law
If two objects interact, the force exerted by object 1 on object 2 is equal in magnitude and opposite in direction to the force exerted by object 2 on object 1.
Particle in Equilibrium model
If the acceleration of an object is zero, the net force on it is zero: ΣF⃗ = 0.
Particle Under a Net Force model
If an object accelerates, use Newton's second law: ΣF⃗ = ma⃗.
4.7 Applying Newton's Laws — Worked Examples
Example — Block and pulley system
Block S (mass M = 3.3 kg) slides on a frictionless surface, connected by a cord over a pulley to a hanging block H (mass m = 2.1 kg). Find the acceleration and the tension.
Sliding block: T = Ma
Hanging block: T − mg = −ma
Combining: T = [Mm/(M+m)]g and a = [m/(M+m)]g
Plugging in numbers: a = 3.8 m/s² and T = 13 N
Sanity check (from the slides): Check that dimensions are correct, that a < g, that T < mg (otherwise the acceleration would be upward), and check limiting cases (e.g., g = 0, M = 0, m → ∞).
Example — Puck sliding to a stop (finding μₖ)
A hockey puck on a frozen pond is given an initial speed of 20.0 m/s and slides 115 m before stopping. Find the coefficient of kinetic friction between puck and ice (using the kinematics equation v² = v₀² + 2a(x−x₀) together with Newton's second law and the kinetic-friction formula below).
4.8 Forces of Friction
Why friction matters
Friction is essential for walking, biking, driving, writing, and building — but overcoming friction is also important for efficiency (e.g., in engines).
Two types of friction
Static friction: the opposing force that prevents an object from starting to move; can take any value from 0 up to a maximum; once that maximum is exceeded, the object slides.
Kinetic friction: the opposing force acting on an object already in motion; has one value, generally smaller than the maximum static friction.
Maximum static friction fs,max = μs FN
Kinetic friction fk = μk FN
Important
FN is the magnitude of the normal force (how strongly the surfaces are pushed together). The friction coefficients μs and μk are unitless and must be determined experimentally.
🎯 Exam Focus — Chapter 4
- F⃗net = ma⃗ is the single most-used equation in this course — always identify the body, draw a free-body diagram, and only include forces acting on that body.
- Internal forces between bodies in the same system never appear in the FBD of the whole system.
- Mass is constant everywhere; weight (W = mg) changes with location. Don't confuse them on an exam.
- In pulley/two-block problems, always write Newton's second law separately for each block, then combine equations — see the worked example above.
- fs,max = μsFN gives the maximum possible static friction — actual static friction can be anything from 0 up to this value, matching the applied force, until sliding starts.
This chapter introduces the concept of a "system," work done by a force, kinetic energy and the work–kinetic-energy theorem, gravitational potential energy, and conservation of energy.
5.1 System and Energy
Definition — System
A system is a small portion of the Universe we choose to consider (ignoring the rest). It may be a single object, a collection of objects, or a region of space, and may change in size/shape over time.
Definition — Energy
Energy is a scalar quantity assigned to an object or system; it can be changed from one form to another, and is conserved in a closed system — the total energy of all types is always the same.
5.2 Work Done by a Constant Force
Definition — Work
The work W done on a system by a constant force is the product of the force magnitude F, the displacement magnitude r of the point of application, and cos θ, where θ is the angle between force and displacement.
Work (constant force) W = F·r·cos θ
Watch out
A force does no work if the point of application does not move, or if the force is perpendicular to the displacement (cos 90° = 0) — this is why the normal force and gravity do no work on an object sliding horizontally.
Sign of work
Work is positive when the force's projection onto the displacement is in the same direction as the displacement; negative when opposite. Work is a scalar. Unit: the joule (J); 1 J = 1 N·m = 1 kg·m²/s².
Work as a dot product
Since work involves the angle between two vectors, it can be written using the scalar (dot) product: W = F⃗·d⃗ = Fd cos θ.
Example — Pulling a vacuum cleaner
A force of magnitude F = 50.0 N is applied at 30.0° to the horizontal, displacing the vacuum cleaner 3.00 m to the right. (Work is computed directly from W = Fd cos θ using these values.)
5.3 Kinetic Energy and the Work–Kinetic-Energy Theorem
Definition — Kinetic Energy
The faster an object moves, the greater its kinetic energy; it is zero for a stationary object. For v well below the speed of light:
Kinetic energy K = ½mv²
Unit of kinetic energy: joule (J), where 1 joule = 1 J = 1 kg·m²/s².
Work–Kinetic-Energy Theorem
When work is done on a system and the only change is in the speeds of its members, the net work done equals the change in kinetic energy:
Work–KE theorem W = ΔK (i.e., final KE = initial KE + net work)
This theorem holds for both positive and negative net work.
Example — Adding/removing kinetic energy
If a particle's kinetic energy is initially 5 J:
- Net transfer of +2 J to the particle → Final KE = 5+2 = 7 J
- Net transfer of −2 J from the particle → Final KE = 5−2 = 3 J
Example — Block pulled by constant force (Figure 7.14)
A 6.0 kg block, initially at rest, is pulled by a constant horizontal force of 12 N across a frictionless surface through 3.0 m. The net external force is the 12 N force; using the work–KE theorem with initial KE = 0, the work W = Fd is set equal to ΔK = ½mv
f², which is then solved for the final speed v
f.
5.4 Potential Energy of a System
Definition
Potential energy U is energy associated with the configuration of a system of objects that exert forces on each other. Gravitational potential energy accounts for the kinetic energy gained during a fall.
Gravitational potential energy Ug = mgy
The work done in lifting an object slowly through a vertical displacement must appear as an increase in the system's energy — stored as gravitational potential energy. Unit: joule (J).
Example — Trophy falling
Choosing floor level as y = 0, the gravitational potential energy of the trophy–Earth system is calculated just before release and again when the trophy reaches the athlete's foot, using U
g = mgy at each point; the change in U
g is the difference between these two values. The slides also note the calculation can be repeated using the top of the athlete's head as the origin instead.
5.5 Isolated Systems and Conservation of Energy
| Non-isolated system | Isolated system |
|---|
| Energy can cross the system boundary in a variety of ways; total energy of the system changes. | Energy does not cross the boundary; total energy of the system is constant. |
Conservation of energy
Energy cannot be created nor destroyed — if the total energy of a system changes, it is only because energy crossed the system's boundary by some transfer method.
Isolated system, no non-conservative forces ΔEmech = 0 (Emech = K + U)
Equivalent form Kf + Uf = Ki + Ui
Watch out
This form (Kf+Uf=Ki+Ui) applies only to a system in which conservative forces act. If non-conservative forces act (like friction), some energy is transformed into internal energy, and the more general statement ΔEsystem = 0 must be used instead, where Esystem includes all kinetic, potential, and internal energies.
5.6 Problem-Solving Strategy (Conservation of Mechanical Energy)
Categorize: Define the system; check whether energy transfers occur across its boundary (if so, use the non-isolated system model); check for non-conservative forces (if none, use conservation of mechanical energy).
Analyze: Choose initial/final configurations; identify the zero-configuration for gravitational (and elastic, if any) potential energy; write expressions for total initial and total final mechanical energy and set them equal.
Example — Ball dropped from height h
A ball of mass m is dropped from height h. Find its speed at height y above the ground.
Conceptualize/Categorize: system = ball + Earth, isolated, only gravity (conservative) acts.
Kf + Ugf = Ki + Ugi, with Ki = 0 (ball is dropped)
½mvf² + mgy = 0 + mgh → vf = √(2g(h − y))
This result is consistent with the particle-under-constant-acceleration model for a falling object (Chapter 1).
🎯 Exam Focus — Chapter 5
- W = Fd cos θ — always identify the angle between the force and the direction of motion; forces perpendicular to displacement do zero work.
- The work–KE theorem (W = ΔK) is often the fastest route to a final speed — you don't always need kinematics equations.
- Kf+Uf=Ki+Ui only holds with NO non-conservative forces (like friction) — this is a very common exam trap.
- Choosing the y = 0 reference point for gravitational PE is a free choice — but you must stay consistent throughout one problem.
- Energy is a scalar — no components, no direction, just add/subtract algebraically.
This chapter covers linear momentum, impulse, conservation of momentum, the three types of collisions, and the center of mass.
6.1 Linear Momentum
Definition
The linear momentum of a particle (mass m, velocity v⃗) is the product of mass and velocity. It is a vector quantity, in the same direction as velocity. SI unit: kg·m/s.
Linear momentum p⃗ = mv⃗ (components: px=mvx, py=mvy, pz=mvz)
Newton's Second Law in terms of momentum
The time rate of change of a particle's momentum equals the net force acting on it — this is the more general form of Newton's second law, and it also allows for mass changes.
Newton's second law (momentum form) F⃗net = dp⃗/dt
6.2 Conservation of Momentum
Conservation of Momentum
The total momentum of any isolated physical system remains constant. For a two-particle isolated system, the total momentum before an interaction equals the total momentum after.
Conservation of momentum (isolated system of two particles) p⃗1i + p⃗2i = p⃗1f + p⃗2f
Component form p1ix+p2ix = p1fx+p2fx, p1iy+p2iy = p1fy+p2fy, p1iz+p2iz = p1fz+p2fz
This is the law of conservation of linear momentum; it applies to systems with any number of particles.
Example — The archer problem
An archer stands on frictionless ice. The system (archer + bow, and arrow) is isolated in momentum along the x-direction (no external horizontal forces), even though it is not isolated in the y-direction (gravity and the normal force act there).
Total momentum before releasing the arrow = 0
Total momentum after = m₁v₁f + m₂v₂f = 0
With archer mass m₁ = 60 kg and arrow mass m₂ = 0.030 kg, solving for the archer's recoil velocity shows it is negative (opposite the arrow's direction) and much smaller in magnitude, since the archer's mass is much larger.
6.3 Impulse and Momentum
Definition — Impulse
Starting from F⃗net = dp⃗/dt, integrating over a time interval gives the change in momentum. This integral of force over time is called the impulse, J⃗, of the force acting on the object.
Impulse–momentum theorem Δp⃗ = J⃗ = ∫F⃗ dt
Important
Impulse is a vector. Its magnitude equals the area under the force–time curve (the force may vary with time). Impulse is not a property of the particle itself, but a measure of the change in its momentum. This form is equivalent to Newton's second law, and is the most general statement of conservation of momentum — it applies to non-isolated systems.
Impulse approximation
When one force acts on a particle for a very short time but is much greater than any other force present (as in most collisions), that force is called the impulsive force, and the particle is assumed to move very little during the collision.
Example — Crash test
A 1500 kg car collides with a wall; the collision lasts 0.150 s.
(a) The impulse on the car and the average net force are found from J⃗ = Δp⃗ = m(v⃗f − v⃗i), then average force = J⃗/Δt.
(b) If the car does not rebound (final velocity = 0) but the same time interval is used, this represents a smaller net force than the rebounding case, because the change in momentum is smaller.
6.4 Types of Collisions
| Elastic collision | Inelastic collision | Perfectly inelastic collision |
|---|
| Both momentum and kinetic energy are conserved. No transformation of KE into other energy forms. |
Momentum is conserved, but kinetic energy is not — some KE is lost. |
A special case of inelastic collision — the objects stick together and move with the same final velocity. |
Momentum is conserved in all collision types; kinetic energy is conserved only in elastic collisions.
Elastic collision — momentum m₁v⃗₁ᵢ + m₂v⃗₂ᵢ = m₁v⃗₁f + m₂v⃗₂f
Elastic collision — kinetic energy ½m₁v₁ᵢ² + ½m₂v₂ᵢ² = ½m₁v₁f² + ½m₂v₂f²
Perfectly inelastic collision m₁v⃗₁ᵢ + m₂v⃗₂ᵢ = (m₁+m₂)v⃗f
Example — Newton's cradle (five balls)
If one ball strikes an identical row of balls, could two balls exit from the other side? Checking momentum before/after shows it is conserved either way — but checking
kinetic energy before and after shows KE would NOT be conserved if two balls left at half the speed each, which is inconsistent with an elastic collision.
Conclusion: it is not possible for two balls to exit.
6.5 Two-Dimensional Collisions
Momentum is conserved in all directions independently. If particle 1 moves at v⃗₁ᵢ and particle 2 is initially at rest: initial x-momentum = m₁v₁ᵢ, initial y-momentum = 0. After the collision:
x: m₁v₁f cos α + m₂v₂f cos β y: m₁v₁f sin α − m₂v₂f sin β
(the negative sign accounts for a downward velocity component). If the collision is elastic, the kinetic-energy equation provides an additional condition. This scenario is called a glancing collision.
Example 9.8 — Car–truck collision (perfectly inelastic, 2D)
A 1500 kg car traveling east at 25.0 m/s collides with a 2500 kg truck traveling north at 20.0 m/s; they stick together. Applying the isolated-system (momentum) model separately in the x- and y-directions, then dividing the y-equation by the x-equation, gives the direction of the final velocity; substituting back gives its magnitude.
6.6 The Center of Mass
Definition
The center of mass is the special point in a system where all its mass can be considered concentrated — the system moves as if an external force were applied there, independent of rotation, vibration, or deformation.
Center of mass coordinates xCM = Σmixi / M, yCM = Σmiyi / M, zCM = Σmizi / M
where M is the total mass of the system. For a system of particles, the center of mass can also be located by a position vector r⃗CM using the position vector r⃗i of each particle.
Example — Two-body center of mass
Two bodies of masses 5 kg and 15 kg are located at (1,0) and (0,1) respectively. Their center of mass is found by substituting these masses and coordinates directly into the x
CM, y
CM formulas above.
Practical method
For an irregular object, hang it from two or more points, draw the vertical extension of each suspension line — the center of mass is where these lines intersect.
🎯 Exam Focus — Chapter 6
- Momentum is always conserved in an isolated system, in every type of collision. Kinetic energy is conserved ONLY in elastic collisions.
- In perfectly inelastic collisions, objects share a common final velocity — this is the key simplifying condition.
- For 2D collisions, always conserve momentum in the x- and y-directions separately.
- Impulse (J = ∫Fdt = Δp) is useful whenever force is not given directly, but a time interval and velocity change are — very common in crash/collision problems.
- Don't confuse center of mass (a point in space, depends on mass distribution) with center of gravity — for this course they can be treated the same near Earth's surface.
The same laws of motion apply to rotation, but new quantities are needed: angular position, velocity, acceleration, torque, and rotational inertia.
7.1 Rotational Variables
Definition — Angular position
For a rigid body rotating about a fixed axis, a point P at distance r from the axis has angular position θ, measured counterclockwise from a fixed reference line. As the particle moves through angle θ, it sweeps an arc length s, related to r by s = rθ. Angle is measured in radians (dimensionless); θ is not reset to zero after a full rotation.
Angular displacement
Δθ = θf − θi — the angle through which the object rotates during a time interval.
Average angular velocity ωavg = (θ₂−θ₁)/(t₂−t₁) = Δθ/Δt
Instantaneous angular velocity ω = limΔt→0 Δθ/Δt = dθ/dt
If the body is rigid, these equations hold for every point on the body. The magnitude of angular velocity is called angular speed; it is positive if θ is increasing (counterclockwise) and negative if decreasing (clockwise).
Average angular acceleration αavg = (ωf−ωi)/(tf−ti) = Δω/Δt
Instantaneous angular acceleration α = limΔt→0 Δω/Δt = dω/dt
Units & direction
Angular acceleration units: rad/s² (or s⁻², since radians are dimensionless). It is positive if a counterclockwise rotation is speeding up, and negative if a clockwise rotation is slowing down. Strictly, ω and α are magnitudes of vectors whose direction is given by the right-hand rule.
7.2 Correspondence Between Linear and Rotational Quantities
| Linear | Type | Rotational | Relation |
|---|
| x | displacement | θ | x = rθ |
| v | velocity | ω | v = rω |
| atan | acceleration | α | atan = rα |
7.3 Rigid Object Under Constant Angular Acceleration
The equations of motion for constant angular acceleration are the same as for linear motion, with angular quantities substituted for linear ones.
Example — Wheel with constant angular acceleration
A wheel rotates with constant angular acceleration α = 3.50 rad/s². If ω
i = 2.00 rad/s at t
i=0, find the angular displacement after t = 2.00 s.
θf − θi = ωit + ½αt²
Substituting the known values at t = 2 s gives the angular displacement.
7.4 Torque
Definition
To start an object rotating, a force is needed — but the position and direction of the force matter too. The perpendicular distance from the rotation axis to the line along which the force acts is the lever arm. A longer lever arm makes rotating an object easier (e.g., a longer wrench, or a tire iron).
Torque τ = rF sin θ
Direction of torque is found using the right-hand rule. SI unit: N·m (kept as N·m, not converted to joules, even though the dimensions look similar — torque is a different physical quantity from work/energy).
Torque and angular acceleration τ = Iα
The torque causes angular acceleration; the constant of proportionality is the object's moment of inertia, I (analogous to mass in F=ma).
Torque and angular acceleration — derivation
For a particle of mass m rotating in a circle of radius r under tangential force Ft:
Ft = mat
The torque produced is τ = rFt = r(mat) = r(mrα) = (mr²)α. Since mr² is the moment of inertia of a single particle, this shows τ = Iα — torque is directly proportional to angular acceleration, with I as the constant of proportionality. For an extended object (many particles, all sharing the same α), the same relation holds using the object's total moment of inertia.
Example 10.4 — Uniform rod released from horizontal (Figure 10.12)
A uniform rod of length L and mass M is pivoted (frictionless) at one end and released from rest in the horizontal position. Find the initial angular acceleration of the rod and the initial translational acceleration of its right end.
Net external torque about the pivot, due to gravity acting at the rod's center of mass (distance L/2 from the pivot):
Στext = Mg(L/2)
Using τ = Iα with the moment of inertia of a rod about one end, I = ⅓ML² (from the standard moment-of-inertia table referenced in the slides):
α = Στext/I = Mg(L/2) / (⅓ML²) = 3g/2L
Using at = rα with r = L (the right end of the rod):
at = Lα = (3/2)g
🎯 Exam Focus — Chapter 7
- The rotational kinematics equations are a direct "translation" of the linear ones — memorize the correspondence table (x↔θ, v↔ω, a↔α) instead of two separate sets of formulas.
- τ = rF sin θ — the angle θ here is between r and F; torque is maximized when the force is applied perpendicular to r (θ=90°).
- τ = Iα is the rotational analog of F = ma; I (moment of inertia) plays the role mass plays in linear motion — it depends on how mass is distributed relative to the axis, not just total mass.
- Don't drop factors of ½ or ⅓ that come from specific moment-of-inertia formulas (e.g., I=⅓ML² for a rod about its end) — always check which shape/axis the formula in your table applies to.
- Torque units are N·m, never joules, even though both reduce to kg·m²/s² dimensionally.
This chapter introduces electric charge, how objects become charged, Coulomb's law for the force between point charges, the electric field, and electric field lines.
8.1 Properties of Electric Charges
Static electricity
Rubbing certain materials (hard rubber, glass, plastic) with a cloth causes them to display static electricity — the object becomes "charged," possessing a net electric charge. There are two types: positive and negative.
Basic rule of charge interaction
Unlike charges attract; like charges repel. (Demonstrated with charged rulers and glass rods repelling their own kind, but attracting the opposite kind.)
Conservation of electric charge
Electric charge is always conserved in an isolated system — rubbing does not create charge, it only transfers it: one object gains a certain amount of negative charge while the other gains an equal amount of positive charge.
| Material type | Behavior |
|---|
| Electrical conductors | Some electrons are free and can move relatively freely through the material. |
| Electrical insulators | All electrons are bound to atoms and cannot move freely through the material. |
| Semiconductors | Intermediate between conductors and insulators (e.g., silicon, germanium — used in computer chips). |
8.2 Charging Objects by Induction
When a positively charged object is brought close to (but not touching) a neutral metal rod, free electrons move toward the external positive charge, leaving a positive charge at the far end. No net charge is created — charge is merely separated. This is called charging by induction.
8.3 Coulomb's Law
Definition
The electrical behavior of electrons and protons is well described by modeling them as point charges. The magnitude of the electrical force between two point charges q₁ and q₂ separated by distance r is given by Coulomb's law:
Coulomb's Law (magnitude) F = ke |q₁||q₂| / r²
where ke is the Coulomb constant. The slides use the value ke = 8.988 × 10⁹ N·m²/C² (as used directly in the worked charged-spheres example below), and state that ke can also be written in terms of the permittivity of free space, ε₀.
Vector form of Coulomb's Law F⃗₁₂ = ke (q₁q₂/r²) r̂₁₂
Newton's third law applies
The electric force exerted by charge 1 on charge 2 is equal in magnitude and opposite in direction to the force exerted by charge 2 on charge 1: F⃗₂₁ = −F⃗₁₂.
Charge and mass table (Table 22.1)
| Particle | Charge (C) | Mass (kg) |
|---|
| Electron (e) | −1.602 176 5 × 10⁻¹⁹ | 9.109 4 × 10⁻³¹ |
| Proton (p) | +1.602 176 5 × 10⁻¹⁹ | 1.672 62 × 10⁻²⁷ |
| Neutron (n) | 0 | 1.674 93 × 10⁻²⁷ |
Superposition of electrostatic forces
If several charges act on one particle, the net (resultant) force is the vector sum (not scalar sum) of the individual forces from each other charge.
Example — Electron and proton in a hydrogen atom
Using Coulomb's law with the electron and proton charge magnitudes (equal), and the given average orbital distance, the force is attractive (opposite-sign charges), directed toward the proton.
Example 22.2 — Three charges on a right triangle (Figure 22.8)
q₁ = q₃ = 5.00 μC, q₂ = −2.00 μC, a = 0.100 m. The force from q₁ on q₃ is repulsive (both positive), at 45.0° to the x-axis. Coulomb's law gives the magnitude of each pairwise force using the absolute values of the charges; x- and y-components of each force are found and summed to get the resultant force on q₃, then expressed in unit-vector form.
Example 22.4 — Two charged spheres in equilibrium (Figure 22.10)
Two identical small spheres, each of mass 3.00 × 10⁻² kg, hang from strings of length L = 0.150 m, each making angle θ = 5.00° with the vertical. Find the charge on each sphere.
From the particle-in-equilibrium model (net force = 0 on each sphere):
(1) T sin θ − Fe = 0 → T sin θ = Fe
(2) T cos θ − mg = 0 → T cos θ = mg
Dividing (1) by (2):
(3) tan θ = Fe/mg → Fe = mg tan θ
From the geometry of the right triangle (half-separation a, string length L):
(4) sin θ = a/L → a = L sin θ
Solving Coulomb's law for the charge and substituting (3) and (4):
|q| = √[ Fe(2a)² / ke ] = √[ mg tan θ (2L sin θ)² / ke ]
Substituting numbers: m=3.00×10⁻² kg, g=9.80 m/s², θ=5.00°, L=0.150 m, ke=8.988×10⁹ N·m²/C²:
|q| = 4.42 × 10⁻⁸ C
8.4 Analysis Model: Particle in a Field (Electric)
Definition — Electric Field
An electric field exists in the region of space around a charged object (the source charge). The electric field vector E⃗ at a point P is the electric force acting on a positive test charge placed at that point, divided by the magnitude of the test charge.
Electric field E⃗ = F⃗e / q₀
Force on a charge placed in a field F⃗e = qE⃗
SI unit of E: newtons per coulomb (N/C). The direction of E⃗ is parallel to the force on a positive test charge. If q is negative, the force on it — and hence the field's effect — points toward the source charge.
Electric field of a single point charge E = ke |q| / r²
Example 22.5 — Water droplet suspended in an electric field
A water droplet of mass 3.00 × 10⁻¹² kg is suspended at rest by an atmospheric electric field of magnitude 6.00 × 10³ N/C pointing vertically downward.
Newton's second law (particle in equilibrium), vertical direction:
(1) ΣFy = 0 → Fe − Fg = 0
Substituting Fe = qE (with E negative, since it points down) and Fg = mg:
q(−E) − mg = 0 → q = −mg/E
Substituting numbers:
q = −(3.00×10⁻¹² kg)(9.80 m/s²) / (6.00×10³ N/C) = −4.90 × 10⁻¹⁵ C
Example 22.6 — Electric field due to two charges (Figure, on the x-axis)
Charges q₁ and q₂ sit on the x-axis at distances a and b from the origin. The total field at point P is the vector sum E⃗ = E⃗₁ + E⃗₂.
E₁ = ke|q₁| / (a²+y²) E₂ = ke|q₂| / (b²+y²)
In unit-vector form (φ, θ are the angles each field vector makes, from the given geometry):
E⃗₁ = ke[|q₁|/(a²+y²)] cos φ î + ke[|q₁|/(a²+y²)] sin φ ĵ
E⃗₂ = ke[|q₂|/(b²+y²)] cos θ î − ke[|q₂|/(b²+y²)] sin θ ĵ
The components of the net field are found by adding the corresponding x- and y-components of E⃗₁ and E⃗₂.
8.5 Electric Field Lines
Rules for drawing electric field lines
- The field vector E⃗ is tangent to the field line at every point.
- Lines are close together where the field is strong, and far apart where it is weak (line density indicates field strength).
- Lines must begin on a positive charge and terminate on a negative charge (or go to infinity for an isolated charge).
- The number of lines leaving a positive charge (or entering a negative one) is proportional to the magnitude of the charge.
- No two field lines can ever cross, even with multiple charges present.
Field lines point outward from a positive charge and inward toward a negative one; lines get closer together near a charge, showing the field grows stronger closer to the source.
🎯 Exam Focus — Chapter 8
- Coulomb's law and the electric-field formula have the same r² denominator and the same constant ke — E is just F per unit charge (E=F/q₀), so if you know one you can derive the other.
- Always use the absolute value of the charges to compute the magnitude of a Coulomb force, then reason separately about attraction vs. repulsion from the signs.
- When multiple charges act on one charge, add forces (or fields) as vectors, not as magnitudes — resolve into x/y components first.
- Electric field direction: away from positive source charges, toward negative source charges — remember this is defined using a positive test charge.
- Field lines never cross, and their density (closeness) directly indicates field strength — a common conceptual exam question.
This chapter extends the electric-field concept to continuous distributions of charge, introduces electric flux, and presents Gauss's law as a more general, elegant alternative to Coulomb's law.
9.1 Electric Field of a Continuous Charge Distribution
Definition
When charge is continuously distributed along a line, over a surface, or through a volume, the field is found by: (1) dividing the distribution into small elements each carrying a small charge, (2) using Coulomb's law (vector form) to find the field due to one element at point P, and (3) summing the contributions of all elements (superposition).
Field due to one charge element dE⃗ = ke (dq/r²) r̂
Total field (sum over all elements) E⃗ ≈ Σᵢ ke (Δqᵢ/rᵢ²) r̂ᵢ
where r is the distance from a charge element to point P, and r̂ is a unit vector directed from the element toward P.
Charge density definitions
Volume charge density ρ = Q / V (units: C/m³)
Surface charge density σ = Q / A (units: C/m²)
Linear charge density λ = Q / ℓ (units: C/m)
These apply when a charge Q is uniformly distributed throughout a volume V, over a surface of area A, or along a length ℓ, respectively.
9.2 Electric Flux
Definition
Consider a uniform electric field penetrating a rectangular surface of area A, with the plane of the surface perpendicular to the field. The product of E and the perpendicular area is called the electric flux.
Electric flux (field perpendicular to surface) ΦE = EA
Electric flux (field at angle θ to the surface normal) ΦE = EA cos θ
where θ is the angle between the field direction and the line perpendicular (normal) to the surface. The flux is maximum (ΦE=EA) when the surface is perpendicular to the field (normal parallel to the field, θ=0).
Example — Flux through a tilted surface
A uniform field E = 400 N/C is incident on a surface of area A = 10 m² at an angle of 30° to the surface.
ΦE = EA cos θ = 400 × 10 × cos 30° = 3464 N·m²/C
9.3 Gauss's Law
Definition
Gauss's law relates the total electric flux through a closed surface of any shape to the net charge enclosed within that surface. For any closed surface, divided into tiny area elements (each small enough that E is essentially constant across it), the total flux is the sum of the flux through each tiny element — and this total flux is proportional to the net charge enclosed.
Gauss's Law ΦE = qenc / ε₀
where 1/ε₀ is the constant of proportionality and qenc is the net charge enclosed within the surface.
Important
Gauss's law is a more general and elegant form of Coulomb's law. Both Coulomb's law and Gauss's law can be used to find the electric field due to a given (static) charge distribution — but Gauss's law is especially useful when the charge distribution is simple and symmetrical.
🎯 Exam Focus — Chapter 9
- Know the three charge-density definitions (ρ=Q/V, σ=Q/A, λ=Q/ℓ) and be able to tell which applies from the geometry described (a volume, a surface, or a line/rod of charge).
- Electric flux ΦE=EA cos θ — θ is measured from the surface's normal (perpendicular direction), not from the surface itself; this is a very common source of error.
- Flux is maximum when the field is perpendicular to the surface (parallel to the normal), and zero when the field runs parallel to the surface (perpendicular to the normal).
- Gauss's law (ΦE=qenc/ε₀) only involves the charge enclosed by the surface — charges outside the chosen (Gaussian) surface contribute zero net flux through it.
- Gauss's law is a shortcut for highly symmetric charge distributions (spheres, cylinders, planes) — for irregular distributions, direct integration (Coulomb's law summed over elements) is used instead.
Most Important Definitions
- Speed (scalar) vs. velocity (vector, has direction).
- Acceleration: rate of change of velocity — not the same as deceleration.
- Vector (magnitude + direction) vs. scalar (magnitude only).
- Force: a push or pull that causes acceleration; unit is the newton (N).
- Mass (resistance to acceleration, constant) vs. weight (W=mg, depends on location).
- Work: W = Fd cos θ — force times displacement times the cosine of the angle between them.
- Kinetic energy: energy of motion, K=½mv².
- Potential energy: energy of configuration, Ug=mgy (gravitational).
- Linear momentum: p⃗=mv⃗ — a vector, conserved in every isolated system/collision.
- Impulse: J⃗=Δp⃗=∫F⃗dt — the "kick" a force gives over time.
- Torque: τ=rF sin θ — the rotational analog of force.
- Moment of inertia (I): the rotational analog of mass — resistance to angular acceleration.
- Electric field: E⃗=F⃗/q₀ — force per unit (positive test) charge.
- Electric flux: ΦE=EA cos θ — a measure of field lines passing through a surface.
Most Important Laws & Formulas
| Topic | Formula |
|---|
| Kinematics (constant a) | x=x₀+v₀t+½at² | v=v₀+at | v²=v₀²+2a(x−x₀) |
| Projectile range | R = (v₀²/g) sin 2θ₀ |
| Centripetal acceleration | a = v²/r |
| Dot product | a⃗·b⃗ = ab cos φ = aₓbₓ+aᵧbᵧ+a_zb_z |
| Cross product | |a⃗×b⃗| = ab sin φ |
| Newton's Second Law | F⃗net = ma⃗ |
| Friction | fs,max=μsFN | fk=μkFN |
| Work–KE theorem | Wnet = ΔK |
| Conservation of mechanical energy | Ki+Ui = Kf+Uf (conservative forces only) |
| Conservation of momentum | p⃗1i+p⃗2i = p⃗1f+p⃗2f |
| Elastic collision (2nd condition) | ½m₁v₁ᵢ²+½m₂v₂ᵢ² = ½m₁v₁f²+½m₂v₂f² |
| Rotational kinematics | ωf=ωi+αt | θf=θi+ωit+½αt² |
| Torque | τ = rF sin θ = Iα |
| Coulomb's Law | F = ke|q₁||q₂|/r² |
| Electric field of a point charge | E = ke|q|/r² |
| Gauss's Law | ΦE = qenc/ε₀ |
Most Important Comparisons
| Compare | Key difference |
|---|
| Negative acceleration vs. deceleration | Negative accel. = acceleration in the negative coordinate direction. Deceleration = acceleration opposite to velocity (object slowing down). Not always the same thing. |
| Dot product vs. cross product | Dot → scalar, uses cos φ, commutative. Cross → vector (perpendicular to both), uses sin φ, anti-commutative. |
| Mass vs. weight | Mass is constant matter content; weight (mg) depends on gravity/location. |
| Static vs. kinetic friction | Static friction resists the start of motion (variable, up to fs,max); kinetic friction acts during motion (single value fk, generally smaller). |
| Elastic vs. inelastic vs. perfectly inelastic collisions | Elastic: momentum + KE conserved. Inelastic: only momentum conserved. Perfectly inelastic: objects stick together, share final velocity. |
| Isolated vs. non-isolated system (energy/momentum) | Isolated: no energy/momentum crosses the boundary, total is constant. Non-isolated: energy/momentum can be transferred across the boundary. |
| Linear vs. rotational quantities | x↔θ, v↔ω, atan↔α, m↔I, F↔τ — each rotational equation mirrors a linear one. |
Most Important Key Terms
Vector, scalar, displacement, instantaneous velocity, centripetal acceleration, inertial frame, normal force, tension, coefficient of friction, system (isolated/non-isolated), conservative force, impulse, elastic/inelastic collision, center of mass, angular velocity/acceleration, torque, moment of inertia, point charge, Coulomb constant (ke), electric field, electric flux, permittivity of free space (ε₀).
⚡ Night-Before-the-Exam: The Absolute Essentials
- The 3 kinematics equations (Ch. 1) and their 2D/projectile versions (Ch. 3).
- F⃗net = ma⃗, and how to build a free-body diagram (Ch. 4).
- W = Fd cos θ, K = ½mv², and conservation of mechanical energy (Ch. 5).
- Conservation of momentum, and the difference between elastic/inelastic/perfectly-inelastic collisions (Ch. 6).
- The linear ↔ rotational correspondence table, and τ = Iα (Ch. 7).
- Coulomb's Law F=keq₁q₂/r² and E=keq/r² (Ch. 8).
- Gauss's Law ΦE=qenc/ε₀, and when it's useful (Ch. 9).
- I can state the difference between speed and velocity, and between velocity and acceleration.
- I can derive/use all three constant-acceleration (kinematics) equations, and know which one to pick.
- I understand the difference between negative acceleration and deceleration.
- I can add/subtract vectors graphically and by components.
- I know when to use the dot product vs. the cross product, and can compute both.
- I can split projectile motion into independent horizontal and vertical parts.
- I know the formula for centripetal acceleration and when velocity is constant vs. when it's changing direction.
- I can draw a free-body diagram and apply F⃗net = ma⃗ to a multi-body system (like a pulley problem).
- I know the difference between mass and weight, and between static and kinetic friction.
- I can apply the work–kinetic-energy theorem to find a final speed.
- I know when Ki+Ui=Kf+Uf applies (and when it doesn't — non-conservative forces).
- I can apply conservation of momentum in 1D and 2D collisions.
- I know the difference between elastic, inelastic, and perfectly inelastic collisions.
- I can find the center of mass of a simple system of point masses.
- I know the linear-to-rotational correspondence (x↔θ, v↔ω, a↔α, F↔τ, m↔I).
- I can compute torque and apply τ = Iα.
- I can apply Coulomb's Law to find the force between point charges, including direction.
- I understand the electric field concept (E=F/q₀) and can compute the field of a point charge.
- I understand electric flux and know when Gauss's Law is the most useful tool.
Note on source accuracy: This guide was built strictly from the content of the uploaded course slides (373 slides, Weeks 1–5 and 7–11). Every definition, law, formula, and worked example above was read directly from the slides — nothing was added from outside the material. A very small number of equation boxes in the Electrostatics chapter (the Coulomb's Law and Gauss's Law formula images specifically) did not render in the original PDF export from the source presentation; where this happened, the exact same formula was confirmed as it appears correctly elsewhere in the same slide deck (e.g., the identical Coulomb constant ke=8.988×10⁹ N·m²/C² used in the charged-spheres worked example, and the identical E=keq/r² field formula shown in the two-charges example) rather than taken from any external source.