PHY 103Calculus-based

Mechanics, with the calculus left in.

An interactive module for every topic in PHY 103, in the order of the course: kinematics and Newton's laws through rotation, orbits, oscillations and waves. Every graph is live. Drag it, change it, and check the numbers against the equations.

Start with motion graphs Go to the practice set

The modules follow the week-by-week order of PHY 103 at Princeton University, so you can study along with the course. This is an independent, unofficial study aid. It is not affiliated with, reviewed by, or endorsed by Princeton University or its Physics Department. Use it to study, not on graded work, and follow the course's collaboration rules.

Built by Jiarui Li · Princeton Class of 2030 · a student project© 2026 Jiarui Li. All rights reserved. · Privacy policy · Terms of useLast updated October 23, 2026Spotted a mistake or something confusing?

Strobe photo, one flash every 0.1 s. The dropped ball and the ball launched sideways fall the same distance in every interval: horizontal and vertical motion are independent.

See it in 3D

New to physics? Start here

You don't need to know any physics yet. You do need to be comfortable with algebra; the calculus is explained as it comes up.

How to use a module

Read The intuition: the idea in everyday words.
Watch the picture of what is happening, then the graphs. Do Try this: guess first, then open the answer.
Work the Worked example one step at a time, starting with its free-body diagram.
Read the equations and the explanation.
Answer Check yourself, then do the practice problems.

Look for dotted words: tap them for a quick definition. Every simulation has a Coach that explains whatever control you just moved, and a Reset button.

Five ideas you'll use everywhere

Slope = rate. How steep a graph is tells you how fast something changes.
Area = total. The area under a graph adds up a quantity over time.
Arrows split into parts. Break vectors into x and y and handle each separately.
Forces change motion: ΣF = ma.
Some things never change: energy, momentum and angular momentum are conserved when nothing outside interferes.

Suggested path

Each week: do that week's modules alongside the lectures and problem sets.

Before a midterm or final: drill Which principle?, build a mixed set in Exam review, and clear your mistakes queue.

Rusty on calculus or vectors? The optional Math toolkit at the top of the menu covers derivatives, integrals, dot and cross products, and differential equations from scratch.

Stuck on notation? The Key formulas page in the menu lists the main equations.

Use it to study, not on graded work. Follow your course's rules: don't use this site on homework you submit, in labs, or during quizzes and exams.

Modules by week

How this maps to your course

Each row is one week of PHY 103. Unofficial study aid, not affiliated with Princeton University.

Week	Topic	Modules

Week

Vectors

Mechanics runs on arrows and rates of change. Drag the two arrows to see how components, the dot product and the cross product respond.

The intuition

A vector is a set of directions: "go 3 east and 1.5 north." Adding vectors is following one set of directions, then the next.

The dot product asks "how much do these point the same way?" Pushing a cart along its motion does a lot of work; pushing sideways does none. The cross product asks "how much do these twist?" A wrench turns best when you push perpendicular to it.

A·B = A_xB_x + A_yB_y = AB cos θa number

(A×B)_z = A_xB_y − A_yB_x|A×B| = AB sin θ

Drag the arrow tips

A—

B—

Angle θ—

A · B—

(A × B)_z—

|A + B|—

The shaded parallelogram has area |A × B|. The thick orange segment along B is the shadow of A on B, length A cos θ. A positive cross product means B is counterclockwise from A.

Try this

Predict what will happen first. Then set it up and check.

Drag B until the angle readout says 90°.
What you should see
A · B drops to 0. A force perpendicular to the motion does no work.
Drag B so it points the same way as A.
What you should see
The cross product drops to 0 and the parallelogram flattens. Parallel arrows don't twist anything.
Move B from one side of A to the other.
What you should see
The cross product flips sign: A × B = −(B × A).

Worked example

You pull a sled 20 m across level snow with a rope that makes 30° with the ground. The rope tension is 50 N. How much work do you do?

Try it yourself first, then reveal one step at a time.

Sled. The rope pulls up and forward.

Picture and FBD. The sled moves right. Forces: rope tension T at 30°, weight mg, normal force N, friction f.
Write the vectors. T = (50 cos 30°, 50 sin 30°) = (43.3, 25) N, Δr = (20, 0) m.
Dot product. W = T·Δr = 43.3·20 + 25·0 = 866 J.
Same thing another way. W = Td cos θ = 50·20·cos 30° = 866 J. ✓
Interpret. The vertical part of the pull does no work, because it is perpendicular to the motion. It only reduces N.

Common mistakes

Adding lengths instead of components. |A + B| is usually less than |A| + |B|.
Leaving the calculator in degrees when a formula needs radians.
Forgetting that A · B is a number while A × B is a vector.

The toolkit

Components. A_x = A cos θ, A_y = A sin θ. Add vectors by adding components. Almost every 2D problem becomes two 1D problems this way.
Dot product → work. W = F·Δr. Only the part of the force along the motion counts. Make A and B perpendicular and watch A·B hit zero.
Cross product → twisting. Torque τ = r × F and angular momentum L = r × p. Right-hand rule: fingers along the first vector, curl toward the second, thumb gives the direction.
Units as a lie detector. Check units on every answer. A pendulum's period can only depend on L and g as √(L/g), because that is the only combination with units of seconds.
Small angles. In radians, sin θ ≈ θ and cos θ ≈ 1 − θ²/2. This turns a pendulum into simple harmonic motion in Week 9.

Check yourself: two vectors have A·B < 0. The angle between them is:

WeekOptionalCalculus

Derivatives: how fast things change

A derivative tells you how fast something is changing right now. In physics it turns position into velocity and velocity into acceleration.

The intuition

Check your car's odometer twice, one second apart: the difference is your average speed over that second. Shrink the gap to a hundredth of a second, then a thousandth: the answer settles on one number. That limit is the derivative, your speed at that instant.

On a graph it is the slope of the line that just touches the curve at one point, the tangent line.

Point t

Shrink the gap h

Secant slope Δx/Δt—

Exact derivative—

Difference—

Try this

Predict what will happen first. Then set it up and check.

Shrink the gap h to its smallest value.
What you should see
The secant slope settles onto the tangent slope. That limiting value is the derivative.
Pick sin t and move the point to a peak.
What you should see
At the top of the hill the tangent is flat: derivative 0. For motion, that is a turnaround point.
Pick the exponential.
What you should see
Its slope is always exactly half its height: d(e^(t/2))/dt = ½e^(t/2).

Worked example

A mass on a spring has position x(t) = 5 sin(2t) cm. Find v(t), a(t), and the acceleration at t = 0.5 s.

Try it yourself first, then reveal one step at a time.

A pure math step: no forces needed here. (The spring's FBD is in Simple harmonic motion.)

Identify the inside function. x = 5 sin(u) with u = 2t, so du/dt = 2.
Velocity (chain rule). v = 5 cos(2t)·2 = 10 cos(2t) cm/s.
Acceleration. a = −20 sin(2t) cm/s².
At t = 0.5 s. a = −20 sin(1) = −16.8 cm/s² (radians!).
Check. a = −4x: exactly SHM with ω² = 4. ✓

Common mistakes

Forgetting the chain rule: d/dt sin(ωt) = ω cos(ωt), not cos(ωt).
Expecting a constant to survive. The derivative of x₀ is 0.
Confusing the value of a function with its slope.

Rules you will use

Rule	Math	Physics example
Power	d(tⁿ)/dt = ntⁿ⁻¹	x = 4.9t² gives v = 9.8t
Constant	d(c)/dt = 0	the starting position x₀ drops out of v
Sum and multiple	d(af + bg) = a df + b dg	differentiate term by term
Sine, cosine	d(sin ωt)/dt = ω cos ωt	SHM velocity from position
Exponential	d(e^kt)/dt = ke^kt	drag and damping
Chain rule	df(g(t))/dt = f′(g)·g′(t)	a = v dv/dx
Product rule	d(fg) = f′g + fg′	dp/dt = m dv/dt + v dm/dt for rockets

Check yourself: if x(t) = 4t² (meters), what is the velocity at t = 3 s?

WeekOptionalCalculus

Integrals: adding up

An integral adds up lots of tiny pieces. In physics it turns velocity into distance, force into work, and bits of mass into a moment of inertia.

The intuition

Drive at 30 mph for 2 hours and you go 60 miles: speed × time, the area of a rectangle. If your speed keeps changing, chop the trip into short pieces where the speed is nearly constant, add up the thin rectangles, and make them thinner and thinner.

The total becomes the area under the speed graph. That is the integral.

Integrate up to t

Number of rectangles

Sum of rectangles—

Exact integral—

Error—

Try this

Predict what will happen first. Then set it up and check.

Use just 1 rectangle, then 60.
What you should see
One rectangle is a rough guess; 60 thin ones hug the curve and match the exact area.
Pick 10 − 2t and integrate to t = 8.
What you should see
After t = 5 the velocity is negative: that area counts as negative (moving backward).
Pick sin t and integrate a full cycle (to 6.28).
What you should see
The positive and negative areas cancel: zero net displacement.

Worked example

An object's velocity is v(t) = 3t² − 2t m/s. How far does it move between t = 1 s and t = 3 s?

Try it yourself first, then reveal one step at a time.

A pure math step: no forces needed here.

Set up. Displacement is the area under v: Δx = ∫₁³ (3t² − 2t) dt.
Antiderivative. F(t) = t³ − t².
Plug in the limits. F(3) − F(1) = (27 − 9) − (1 − 1) = 18 m.
Distance or displacement? v(t) = t(3t − 2) is positive for all t > 2/3 s, so it never reverses on this interval: distance = 18 m too.
Check. Average velocity 18 m / 2 s = 9 m/s, between v(1) = 1 and v(3) = 21 m/s. ✓

Common mistakes

Forgetting the limits, or the starting value, when you integrate.
Reading area below the axis as positive. It means motion in the negative direction.
Using ∫v dt for distance when v changes sign. Distance is ∫|v| dt.

Rules you will use

Rule	Math	Physics example
Power	∫tⁿ dt = tⁿ⁺¹/(n + 1)	∫9.8t dt = 4.9t²
Definite integral	∫_a^b f dt = F(b) − F(a)	displacement between two times
Sine, cosine	∫cos ωt dt = (1/ω) sin ωt	position from SHM velocity
Exponential	∫e^kt dt = e^kt/k	distance coasting with drag
1/x	∫dx/x = ln\|x\|	rocket equation
Physics uses	Δx = ∫v dt, W = ∫F dx, I = ∫r² dm	areas everywhere

Check yourself: an object moves at a constant 3 m/s for 4 s. What is the area under its v–t graph?

WeekOptionalCalculus

Dot and cross products, step by step

Two ways to multiply vectors. The dot product gives a number that measures how much two arrows point the same way. The cross product gives a new arrow that measures how much they twist.

The intuition

Dot product: shine a light straight down onto A and look at its shadow along B. Shadow length × length of B = A · B. Same direction: big and positive. Perpendicular: zero. Opposite: negative.

Cross product: its size is the area of the parallelogram the two arrows make, and it points straight out of that parallelogram. Use your right hand: fingers along A, curl them toward B, and your thumb points along A × B.

A

xyz

B

xyz

Try this

Predict what will happen first. Then set it up and check.

Make B point the same way as A.
What you should see
A × B becomes (0, 0, 0): parallel vectors don't twist anything.
Make them perpendicular.
What you should see
A · B = 0, and |A × B| = |A||B|, its maximum.
Swap A and B.
What you should see
The cross product flips direction; the dot product stays the same.

Worked example

A force F = (0, 3, 4) N acts at position r = (2, 1, 0) m from a pivot. Find the torque τ = r × F and its size.

Try it yourself first, then reveal one step at a time.

Force applied at r (drawn in the plane of the page for simplicity).

Write the pattern. r × F = (r_yF_z − r_zF_y, r_zF_x − r_xF_z, r_xF_y − r_yF_x).
x component. 1·4 − 0·3 = 4.
y component. 0·0 − 2·4 = −8.
z component. 2·3 − 1·0 = 6. So τ = (4, −8, 6) N·m.
Size. |τ| = √(16 + 64 + 36) = 10.8 N·m.
Check. τ must be perpendicular to r: r · τ = 2·4 + 1·(−8) + 0 = 0. ✓

Common mistakes

Getting the order wrong. A × B = −(B × A).
Dropping the minus sign in the middle (y) term of the cross product.
Treating A · B as a vector. It is just a number.

Unit vectors

Dot products	Cross products (cyclic order x → y → z → x)
î·î = ĵ·ĵ = k̂·k̂ = 1	î × ĵ = k̂, ĵ × k̂ = î, k̂ × î = ĵ
î·ĵ = ĵ·k̂ = k̂·î = 0	Backward order gives a minus sign: ĵ × î = −k̂. And î × î = 0.

Check yourself: what is î × ĵ?

WeekOptionalCalculus

Differential equations you will meet

Newton's second law is a differential equation: it connects a quantity to how fast it changes. Only a handful of types appear in this course, and they all have known answers.

The intuition

A differential equation is a rule for the slope at every point. Draw a little slope mark everywhere on a grid, drop a starting point, and follow the marks like a leaf floating down a stream. The path you trace is the solution.

Different starting points give different paths, which is why every problem also needs an initial condition.

Starting value v₀

Constant k

Try this

Predict what will happen first. Then set it up and check.

Terminal speed: start at different speeds.
What you should see
Every solution curve heads to the same terminal speed, from above or below.
Decay: double k.
What you should see
The curve drops twice as fast. k sets the time scale, 1/k.
Compare v² drag with linear drag.
What you should see
v² drag kills high speeds fast but has a long, slow tail at low speeds.

Worked example

A puck slows by quadratic drag: dv/dt = −kv², with v(0) = v₀. Find v(t).

Try it yourself first, then reveal one step at a time.

A pure math step. The puck's FBD has one horizontal force, drag, pointing backward.

Separate the variables. All v on one side, all t on the other: dv/v² = −k dt.
Integrate both sides. ∫_v₀^v dv/v² = −k∫₀^t dt gives −1/v + 1/v₀ = −kt.
Solve for v. v = v₀/(1 + kv₀t).
Check t = 0. v = v₀. ✓
Check late times. v → 0 but slowly, like 1/t, unlike the fast exponential decay of linear drag.

Common mistakes

Forgetting the initial condition. It picks which curve you are on.
Mixing variables when separating: get all the v's on one side and all the t's on the other.
Guessing the wrong form for a second-order equation. Use the table.

The equations in this course

Recipe for first-order ones: separate the variables, integrate both sides with limits, then solve for the unknown. For second-order ones, guess the form in the table and check it.

Equation	Where it shows up	Solution
dv/dt = −kv	coasting with linear drag	v = v₀e^−kt
dv/dt = g − kv	falling with linear drag	v = v_T + (v₀ − v_T)e^−kt, v_T = g/k
dv/dt = −kv²	coasting with quadratic drag	v = v₀/(1 + kv₀t)
d²x/dt² = −ω²x	springs, pendulums	x = A cos(ωt + φ)
x″ + γx′ + ω₀²x = 0	damped oscillator	x = Ae^−γt/2 cos(ω_dt + φ)

Check yourself: which function solves dx/dt = −3x?

WeekCalculus

Motion graphs

Position, velocity and acceleration are one function and its derivatives. Slopes take you down the stack. Areas take you back up.

The intuition

Picture driving with a GPS trace (position), a speedometer (velocity) and the push you feel in your seat (acceleration). The speedometer reading is how steep the GPS trace is. The push in your seat is how fast the speedometer needle moves.

Going backward up the chain: the distance you covered is the speedometer reading added up over time, which is the area under the velocity graph.

v = dx/dtslope of x–t

a = dv/dt = d²x/dt²slope of v–t

Δx = ∫ v dtarea under v–t

Δv = ∫ a dtarea under a–t

Time 1.50 s

x (m)—

v (m/s)—

a (m/s²)—

Displacement—

Distance—

—

Drag across the graphs to scrub time. Dashed lines are tangents: the slope of the x tangent is v, the slope of the v tangent is a. Shading is the signed area from t = 0.

Try this

Predict what will happen first. Then set it up and check.

Choose Turnaround and drag time to t = 3 s.
What you should see
v = 0 but a = 2 m/s². Stopped for an instant while still accelerating: zero velocity does not mean zero acceleration.
Keep going to t = 6 s and compare displacement with distance.
What you should see
Displacement is 0 (it ends where it started) but distance is 18 m: 9 m out and 9 m back.
Choose Cubic and find where the x graph goes flat.
What you should see
At t = 2 s and t = 4 s the x graph is flat, and the v graph crosses zero at exactly those times.

Worked example

A particle moves along x with position x(t) = t³ − 6t² + 9t (meters, seconds). (a) When is it at rest? (b) At t = 2.5 s, is it speeding up or slowing down?

Try it yourself first, then reveal one step at a time.

At t = 2.5 s: moving left (v), net force and acceleration to the right.

Picture and FBD. The particle moves on a line. Whatever the forces are, they add to one net force F = ma along x, pointing the same way as a.
Velocity. v = dx/dt = 3t² − 12t + 9 = 3(t − 1)(t − 3).
(a) At rest. v = 0 at t = 1 s and t = 3 s.
Acceleration. a = dv/dt = 6t − 12. At t = 2.5 s: v = 3(1.5)(−0.5) = −2.25 m/s and a = +3 m/s².
(b) Compare signs. v is negative and a is positive: opposite signs, so it is slowing down.
Check. It stops at t = 3 s, so just before that it should indeed be slowing. ✓

Common mistakes

Reading the height of a graph when the question asks about its slope or area.
Thinking negative acceleration always means slowing down. It only does when velocity is positive.
Treating distance and displacement as the same thing when the object turns around.

Reading the graphs

Speeding up or slowing down? Compare the signs of v and a. Same sign: speeding up. Opposite signs: slowing down. A negative acceleration alone tells you nothing about speed.
At rest or turning around where v = 0. It turns around only if v changes sign there.
Displacement vs distance. Displacement is ∫v dt. Distance is ∫|v| dt, so split the integral at every zero of v. Try the Turnaround preset past t = 3 s and watch the two numbers separate.
Position from velocity needs a starting point. x(t) = x₀ + ∫₀^t v(t′) dt′. Free-response questions often give x₀ in a sentence you skim past.

Check yourself: for x = t³ − 9t² + 24t, what is the particle doing at t = 1 s?

Week

Constant acceleration

Integrate a constant a twice and you get every equation you need. Each one leaves out one of the five quantities, so pick the one that skips what you don't know and don't need.

The intuition

If velocity changes at a steady rate, your average velocity is just halfway between start and finish. Distance = average velocity × time. Every equation below is that idea rearranged.

A ball thrown upward loses 9.8 m/s of upward speed every second, stops for an instant at the top, then gains 9.8 m/s of downward speed every second. Same acceleration the whole flight.

v = v₀ + atno Δx

Δx = v₀t + ½at²no v

v² = v₀² + 2aΔxno t

Δx = ½(v₀ + v)tno a

Δx = vt − ½at²no v₀

Equation solver: enter any three, leave two blank

v–t graph of each solution. The shaded area is Δx.

Try this

Predict what will happen first. Then set it up and check.

Load "Ball tossed up". Why are there two answers?
What you should see
The ball passes 5 m above the launch point twice: going up (0.53 s) and coming down (1.92 s).
Load "Braking car" and change a to −10 (brake twice as hard).
What you should see
The stopping distance halves, 62.5 m to 31.25 m. Doubling the speed instead would make it four times longer.
Drop a ball from rest for 2 s: v₀ = 0, a = −9.8, t = 2.
What you should see
Δx = −19.6 m and v = −19.6 m/s. Negative because "up" is positive.

Worked example

A car moving at 25 m/s brakes with its wheels locked and stops in 50 m. Find (a) the deceleration, (b) the stopping time, (c) the coefficient of kinetic friction.

Try it yourself first, then reveal one step at a time.

Car while skidding (moving right).

Picture and FBD. Car moving right. Forces: weight mg down, normal force N up, kinetic friction fₖ backward. Vertically nothing accelerates, so N = mg.
List what you know. v₀ = 25 m/s, v = 0, Δx = 50 m. Unknowns: a and t. The time is unknown, so use the equation without t.
(a) Deceleration. v² = v₀² + 2aΔx gives a = (0 − 25²)/(2·50) = −6.25 m/s².
(b) Time. t = (v − v₀)/a = (0 − 25)/(−6.25) = 4.0 s.
(c) Friction. Horizontal: fₖ = m|a|, and fₖ = μₖN = μₖmg. So μ_k = |a|/g = 6.25/9.8 = 0.64. The mass cancels.
Check. Average speed 12.5 m/s × 4 s = 50 m. ✓

Common mistakes

Using these equations when the acceleration changes.
Mixing sign conventions. If up is positive, a = −9.8 m/s² for anything in free fall.
Saying a = 0 at the top of a throw. Only v is zero there.

Where the equations come from

Start from dv/dt = a with a constant. Integrating once gives v = v₀ + at. Integrating again gives x = x₀ + v₀t + ½at². Eliminate t between them and you get v² = v₀² + 2aΔx.

Free fall is the special case a = −g with up as positive, g = 9.8 m/s². The acceleration is −9.8 m/s² on the way up, at the top and on the way down. At the top, v = 0 but a ≠ 0.

Two answers. The quadratic equations can return two roots. Try "Ball tossed up": a ball passes 5 m above the launch point twice, once rising and once falling. Keep both unless the problem rules one out.

Only when a is constantIf acceleration depends on time, velocity or position, these equations give wrong answers. Drag and changing acceleration covers that case, and exam problems use it often.

Check yourself: a ball is thrown straight up at 20 m/s. Which equation gives the time to reach the top in one step?

WeekCalculus

Drag and changing acceleration

When acceleration changes, go back to the definitions. How you integrate depends on what the acceleration depends on.

The intuition

Drop a ball through air. At first it speeds up like free fall, but drag grows with speed, so the extra speed it picks up each second keeps shrinking. Eventually drag matches gravity and the speed levels off.

The graph of speed approaches its ceiling like a cup filling from a tap that slows as the cup fills: fast at first, then creeping toward full.

Assumption upgrade

Before: acceleration was constant (Constant acceleration)
So we didn't need: anything beyond the five equations
Now: air drag makes the acceleration depend on speed
So we need: calculus: separate variables and integrate

You are given	Method	Result
a(t)	Integrate directly in time	v(t) = v₀ + ∫₀^t a(t′) dt′
a(v)	Separate variables	∫ dv / a(v) = ∫ dt
a(x) or you want v(x)	Chain rule: a = v dv/dx	∫ v dv = ∫ a dx

Try this

Predict what will happen first. Then set it up and check.

Double the mass.
What you should see
Terminal speed doubles (v_T = mg/b). With the same drag constant, a heavier ball falls faster through air.
Start the ball faster than terminal speed.
What you should see
It slows down toward v_T: drag is bigger than gravity, so the acceleration starts negative.
Follow the dashed line from t = 0.
What you should see
It reaches the terminal line at exactly t = τ. If the starting acceleration never weakened, that is when the ball would hit terminal speed.

Worked example

A 2.0 kg ball is dropped from rest. Air resistance is F = −bv with b = 0.50 kg/s. Find (a) the terminal speed and (b) the speed after 4.0 s.

Try it yourself first, then reveal one step at a time.

Ball falling: drag points up, opposite the motion.

Picture and FBD. Take down as positive. Forces: mg down, drag bv up.
Newton's second law. m dv/dt = mg − bv. The acceleration depends on v, so the constant-acceleration equations do not apply.
(a) Terminal speed. Speed stops changing when the forces balance: mg = bv_T, so v_T = 2.0·9.8/0.50 = 39.2 m/s.
Solve the equation. Separating variables gives v = v_T(1 − e^−t/τ) with τ = m/b = 4.0 s.
(b) At 4.0 s. t = τ, so v = 39.2(1 − e⁻¹) = 24.8 m/s.
Check. Less than v_T, and less than the no-air value gt = 39.2 m/s. ✓

Common mistakes

Using v = v₀ + at when a depends on v or t.
Dropping the starting value (the constant of integration).
Thinking a = 0 means stopped. At terminal speed it moves fast with zero acceleration.

Worked case: falling with linear drag

A ball of mass m falls with a drag force −bv. Take down as positive. Newton's second law gives an acceleration that depends on velocity:

a = dv/dt = g − (b/m)v ⟹ ∫_v₀^v dv / (g − v/τ) = ∫₀^t dt ⟹ v(t) = v_T + (v₀ − v_T) e^−t/τ

with time constant τ = m/b and terminal speed v_T = mg/b = gτ.

Linear drag: velocity and acceleration

Mass m

Drag constant b

Initial speed v₀

τ = m/b—

v_T = mg/b—

a at t = 0—

95% of v_T at—

The dashed tangent at t = 0 has slope a(0); it hits the terminal line at exactly t = τ. At t = τ the gap to terminal speed has shrunk to 1/e ≈ 37%. Set v₀ above v_T to see the ball slow down to terminal speed.

Two more patterns to recognize

Coasting with drag, a = −kv: v = v₀e^−kt, and it travels a finite total distance v₀/k even though it never fully stops.
Quadratic drag, a = −kv², asked as "speed after a distance D": use v dv/dx = −kv², so v(x) = v₀e^−kx.

Check yourself: what is the acceleration of a ball that has reached terminal speed?

Week

Projectiles

A projectile is two independent 1D problems that share a clock: constant velocity sideways, free fall vertically. Solve for time in one direction and use it in the other.

The intuition

Horizontal and vertical motion ignore each other. A bullet fired level and a bullet dropped from the same height hit the ground at the same time. Gravity only cares about the up-down part.

So a projectile is a ball rolling at steady speed across the floor while someone else plays an elevator-drop game with its height. Same clock, separate rules.

Simplification for now

We assume: no air resistance
Which lets us: treat horizontal and vertical motion as completely independent, with a_x = 0
This changes in: Drag & changing acceleration, and the drag checkbox in the launcher below

x(t) = (v₀ cos θ) ta_x = 0

y(t) = h + (v₀ sin θ) t − ½gt²a_y = −g

Launcher

Time

Launch speed v₀

Angle θ

Launch height h

Gravity

Air drag, a = −k v

Flight time—

Range—

Max height—

Impact speed—

v_x now—

v_y now—

Try this

Predict what will happen first. Then set it up and check.

Level ground, 30°. Then try 60°.
What you should see
Same range. Angles that add to 90° land in the same spot; 45° goes farthest.
Launch from 20 m up at 34°.
What you should see
About the longest range from this height, well below 45°. Starting high gives extra fall time, so aim flatter.
Stop the ball at the top of its path.
What you should see
v_y = 0 but v_x is unchanged, and the pink acceleration arrow still points straight down.

Worked example

A ball is kicked from level ground at 20 m/s, 37° above horizontal (use sin 37° ≈ 0.60, cos 37° ≈ 0.80). Find the flight time, maximum height and range.

Try it yourself first, then reveal one step at a time.

In flight (no air): gravity is the only force.

Picture and FBD. Only gravity acts, so a_x = 0 and a_y = −g.
Split the launch velocity. v_0x = 20·0.80 = 16 m/s, v_0y = 20·0.60 = 12 m/s.
Maximum height. At the top v_y = 0: H = v_0y²/(2g) = 144/19.6 = 7.35 m.
Flight time. Back to y = 0: t = 2v_0y/g = 24/9.8 = 2.45 s.
Range. R = v_0xt = 16·2.45 = 39.2 m.
Check. Range formula: v₀² sin 2θ/g = 400·0.96/9.8 = 39.2 m. ✓

Common mistakes

Saying the speed is zero at the top. Only the vertical part is.
Using the range formula when launch and landing heights differ.
Putting g into the horizontal equation.

Things exams test

At the top, v_y = 0 but v_x is unchanged, so the speed there is v₀ cos θ, not zero.
The range formula R = v₀² sin 2θ / g only works when launch and landing heights match. Set h = 0 and compare 30° with 60°: same range. Raise h and 45° stops being the best angle.
From a height, solve 0 = h + v_0yt − ½gt² for t and keep the positive root.
With drag, the path is lower, shorter and lopsided: the descent is steeper than the climb. Turn on drag to see it against the dashed no-air path.

Check yourself: at the highest point of a projectile's path (no air), the acceleration is:

Week

Relative motion

Velocities add as vectors when you change reference frames. Write subscripts so the inner ones cancel: boat relative to ground equals boat relative to water plus water relative to ground.

The intuition

Walk forward on a moving walkway and your speed to someone standing beside it is your walking speed plus the walkway's. Velocities stack, and in 2D they stack as arrows.

A boat in a river is a walker on a walkway that slides sideways. The boat's engine controls its motion through the water; the water carries the whole scene downstream.

v_BG = v_BW + v_WG

River crossing, seen from above

v_BW boat relative to water, v_WG current, v_BG what someone on shore sees.

River width

Current speed

Boat speed in water

Heading, upstream of straight

Crossing time—

Downstream drift—

Speed over ground—

Reference frames: who is measuring?

A reference frame is a point of view with its own ruler and clock: "measured from the ground", "measured from the train". The same event looks different from different frames, but the physics works in all of them.

—

Flatcar speed u (relative to ground)

Toss speed, straight up (relative to rider)

Ball velocity, rider frame—

Ball velocity, ground frame—

Acceleration (both frames)—

Quantity	Same in every inertial frame?	Why
Position	No	Depends where you put your origin, and the origins move apart: x′ = x − ut
Velocity	No	v′ = v − u: subtract the frame's velocity
Acceleration	Yes	u is constant, so it drops out when you take d/dt
Forces, mass, ΣF = ma	Yes	Same acceleration and same forces, so Newton's laws hold in every inertial frame
Time	Yes	At everyday speeds clocks agree (relativity changes this near light speed)
Kinetic energy, momentum	No	They depend on velocity. But if momentum or energy is conserved in one inertial frame, it is conserved in all of them

Try this

Predict what will happen first. Then set it up and check.

Point straight across with a strong current.
What you should see
Same crossing time as with a weak current. Only the drift grows.
Aim to land directly opposite.
What you should see
The boat points upstream and crosses more slowly.
Watch which way the boat points.
What you should see
It points upstream of where it actually travels. Heading and direction of motion are different.
Watch the toss from both frames.
What you should see
Ground view: a wide parabola. Switch to the rider's view: straight up and down. Same event, same acceleration arrow, different path.
Stop the flatcar.
What you should see
With u = 0 the two frames are the same frame, so both views match.

Worked example

A plane flies at 200 m/s relative to the air, pointing due north. A wind blows east at 50 m/s. (a) What is its velocity relative to the ground? (b) Which way should it point to travel due north?

Try it yourself first, then reveal one step at a time.

Plane at constant velocity (side view): forces balance.

Picture and FBD. Cruising at constant velocity, so the forces cancel (lift = weight, thrust = drag). The wind changes the velocity we see from the ground, not the forces.
Chain the subscripts. v_PG = v_PA + v_AG = (0, 200) + (50, 0) = (50, 200) m/s.
(a) Size and direction. Speed √(50² + 200²) = 206 m/s, at tan⁻¹(50/200) = 14.0° east of north.
(b) Correct for the wind. Point west of north so the east part cancels: 200 sin φ = 50, φ = 14.5°. Ground speed √(200² − 50²) = 194 m/s.
Check. Correcting for a crosswind always costs some forward speed. ✓

Common mistakes

Adding speeds as numbers instead of adding velocities as arrows.
Mixing up subscripts. Write v_AC = v_AB + v_BC so the inner letters match.
Assuming the shortest path is the fastest.

The two classic questions

Fastest crossing: point straight across. Crossing time is W / v_BW no matter how fast the current is, because only the across component moves you toward the far bank. The current just carries you downstream.
Land directly opposite: aim upstream at sin φ = v_WG / v_BW. This takes longer, t = W / √(v_BW² − v_WG²), and is impossible if the current is faster than the boat.
Inertial frame: one that isn't accelerating (the ground, a train at steady speed, a plane in smooth flight). Newton's laws work as written. A ball dropped in a smooth train falls straight down, exactly as on the platform.
Non-inertial frame: one that speeds up, brakes or turns. Objects seem to get pushed by nothing: coffee sloshes backward when a car accelerates. To use Newton's laws there you add a fictitious force −ma_frame. See Curves & fictitious forces.
Flip the subscripts, flip the sign: v_AB = −v_BA. If the car moves east at 20 m/s relative to you, you move west at 20 m/s relative to the car.
Rain on a moving car: rain falling straight down at 10 m/s looks slanted when you drive at 20 m/s: relative to you it moves backward at 20 m/s and down at 10 m/s, at tan⁻¹(20/10) = 63° from vertical.
Pick the easiest frame. Collisions are simplest in the center-of-mass frame. A river problem is simplest in the water's frame, where the current disappears.

Check yourself: you point straight across and the current doubles. Your crossing time:

WeekCalculus

Circular motion

Even at constant speed, moving in a circle means accelerating, because the direction of velocity keeps changing. Add a change in speed and a second, tangential component appears.

The intuition

Velocity is an arrow, and turning the arrow counts as a change, even at steady speed. Swing a bucket in a circle and your arm constantly pulls it inward to bend its path; let go and it flies off in a straight line, along the tangent.

Tighter circles and higher speeds need a harder inward pull: a_c = v²/r.

Assumption upgrade

Before: motion along straight lines or simple paths
So we didn't need: to think about the direction of velocity changing
Now: velocity keeps turning on a curve
So we need: an inward (centripetal) acceleration v²/r, even at constant speed

a_c = v²/r = ω²rtoward center

a_t = d|v|/dtalong v

|a| = √(a_c² + a_t²)total

Particle on a circle

Radius r

Starting speed v₀

Tangential acceleration a_t

Speed v—

ω = v/r—

Period—

a_c—

|a|—

Faint dots are strobe positions at equal time steps; they spread out as the particle speeds up. Arrows show direction; lengths are scaled per quantity, not to each other.

Try this

Predict what will happen first. Then set it up and check.

Constant speed: set a_t = 0.
What you should see
The total acceleration points straight at the center. Constant speed, still accelerating.
Speed up: set a_t positive.
What you should see
The total acceleration tilts forward and the strobe dots spread apart.
Double the starting speed.
What you should see
The centripetal acceleration becomes 4 times bigger, because a_c ∝ v².

Worked example

A 1200 kg car takes a flat curve of radius 50 m at 15 m/s. Find (a) its centripetal acceleration, (b) the friction force needed, (c) the smallest μₛ that works.

Try it yourself first, then reveal one step at a time.

Seen from behind. The center of the curve is to the left.

Picture and FBD. From behind: mg down, N up, and static friction pointing toward the center. Friction is the only sideways force, so it must do the turning.
(a) Centripetal acceleration. a_c = v²/r = 225/50 = 4.5 m/s² toward the center.
(b) Friction. f = ma_c = 1200·4.5 = 5400 N.
(c) Minimum μₛ. Vertically N = mg. Need f ≤ μ_smg, so μ_s ≥ 4.5/9.8 = 0.46.
Check. At double the speed the car would need four times the friction (a_c ∝ v²).

Common mistakes

"Constant speed means no acceleration." Not on a curve.
Thinking the acceleration points outward.
Using degrees for ω. Angular speed is in radians per second.

Deriving a_c with calculus

For uniform motion, r(t) = r(cos ωt, sin ωt). Differentiate: v = rω(−sin ωt, cos ωt), which is perpendicular to the radius with magnitude rω. Differentiate again: a = −ω²r. It points opposite the position vector, toward the center, with magnitude ω²r = v²/r.

Set a_t to something nonzero and watch the total acceleration tilt away from the center: forward when speeding up, backward when slowing down.

Check yourself: a car doubles its speed around the same curve. Its centripetal acceleration:

Week

Curves and fictitious forces

To follow a curve, something has to pull you toward the center. A banked road, a rope or friction can all do that job.

The intuition

"Centripetal force" is not a new kind of force. It is a job title. Whatever forces are present (friction, the tilt of the road, a rope, gravity) have to add up to mv²/r pointing at the center, or you can't follow the curve.

In a turning car you feel thrown outward. Nothing actually pushes you out: your body is trying to keep going straight while the car turns under you. The "centrifugal force" only appears if you describe things from inside the turning car.

Assumption upgrade

Before: every problem was described from the ground, an inertial frame
So we didn't need: fictitious forces
Now: we sometimes ride along in a turning frame
So we need: either switch back to the ground frame, or add a fictitious centrifugal force mω²r

Car on a banked curve, seen from behind

—

Speed v

Curve radius r

Bank angle θ

Static μ_s

No-friction speed—

Slowest safe—

Fastest safe—

Normal N—

Friction needed—

A 1000 kg car. The pink arrow is the net force, m v²/r toward the center. At the no-friction speed the tilt of the road does the whole job: tan θ = v²/(rg).

Tetherball: a conical pendulum Ride along with the ball (rotating frame)

—

String angle from the pole θ

String length L

Circle radius—

Speed—

Period—

Tension—

Only two real forces act: tension and gravity. Their sum points horizontally at the pole. Tick the box to see the same ball from a frame spinning with it, where it looks stationary and a fictitious outward "centrifugal" force seems to balance the books. Period: T = 2π√(L cos θ/g).

Try this

Predict what will happen first. Then set it up and check.

Drive faster than the no-friction speed.
What you should see
Friction now points down the bank, keeping the car from drifting outward.
Drive slower than it.
What you should see
Friction flips to point up the bank, keeping the car from sliding inward.
Swing the tetherball almost flat.
What you should see
The rope tension climbs to several times mg.
Ride along with the ball.
What you should see
In the rotating view an outward "centrifugal" arrow appears to balance the forces. It has no partner and nothing pushes it.

Worked example

A curve of radius 80 m is banked at 20°. At what speed can a car take it with no friction at all?

Try it yourself first, then reveal one step at a time.

Car from behind on the banked road. Center is to the left.

Picture and FBD. Only two forces: mg down and N perpendicular to the road. N tilts toward the center.
Vertical. No vertical acceleration: N cos 20° = mg.
Horizontal. The horizontal part of N does the turning: N sin 20° = mv²/r.
Divide. tan 20° = v²/(rg), so v = √(80·9.8·tan 20°) = 16.9 m/s.
Check. A flat road (θ = 0) would give v = 0: without banking you need friction at any speed. ✓

Common mistakes

Drawing "centripetal force" as an extra arrow.
Thinking the centrifugal force is real in the ground frame.
Forgetting that the normal force on a banked road has a horizontal part.

How to solve curve problems

Point one axis at the center. Add up the force components along it and set the sum equal to mv²/r. Never draw "centripetal force" as its own arrow.
Flat road: only static friction can point sideways, so the top speed is √(μ_sgr).
Banked road: the tilted normal force helps. At the design speed v² = rg tan θ no friction is needed at all.
Non-inertial frames: in an accelerating or rotating frame, Newton's laws only work if you add fictitious forces, like −ma_frame or the centrifugal force mω²r. They are bookkeeping, not real pushes. In an inertial (non-accelerating) frame they disappear.
Speed-dependent forces like drag lead to the methods in Drag & changing acceleration.

Check yourself: a car rounds a flat, unbanked curve at steady speed. What pulls it toward the center?

Week

Forces and friction

You don't need a force to keep moving; you need a force to change how you move. A block on a ramp is the standard place to practice free-body diagrams, because gravity, the normal force and friction all point in different directions.

The intuition

Objects are lazy: left alone they keep doing whatever they were doing. A force is a nudge that changes velocity, and a heavier object needs a bigger nudge for the same change. That is all of ΣF = ma.

Static friction is a polite bouncer. It pushes back exactly as hard as needed to keep the block still, up to a limit of μ_sN. Past that limit the block breaks free and kinetic friction takes over at a fixed μ_kN.

Assumption upgrade

Before: objects slid on frictionless surfaces or flew through the air (Week 1)
So we didn't need: contact forces along a surface
Now: real surfaces grip
So we need: a friction force at every contact, and a way to decide which way it points: see Which way does friction point?

ΣF = ma2nd law

f_s ≤ μ_sN, f_k = μ_kNfriction

F_AB = −F_BA3rd law

Block on an incline

—

Incline angle θ

Mass m

Static μ_s

Kinetic μ_k

Push along slope, + is up

Normal N—

mg sin θ—

Friction (+ up)—

a (+ up)—

Slips past—

Orange arrows are forces on the block, drawn to scale. The pink arrow is the acceleration. Raise θ slowly with no push and find the angle where it starts to slide.

Try this

Predict what will happen first. Then set it up and check.

With no push, tilt the ramp just past 26.6°.
What you should see
It starts to slide: tan θ = μₛ = 0.5. Change the mass and the angle stays the same.
Push up the slope hard enough to move it.
What you should see
Friction shrank, flipped to point down the slope, and the block slides up once P beats mg sin θ + μₛN (about 32 N).
Remove all friction.
What you should see
Any angle slides, with a = g sin θ. Mass cancels.

Worked example

A 5.0 kg box rests on a 30° ramp. μₛ = 0.40 and μₖ = 0.30. Does it slide? If so, what is its acceleration?

Try it yourself first, then reveal one step at a time.

Box on the ramp. Friction points up the slope.

Picture and FBD. Forces: weight mg straight down, normal force N perpendicular to the ramp, friction f along the ramp (up the slope, opposing the slide).
Tilt the axes. Along the slope: mg sin 30° = 5.0·9.8·0.50 = 24.5 N down. Perpendicular: N = mg cos 30° = 42.4 N.
Does it slide? Maximum static friction μ_sN = 0.40·42.4 = 17.0 N is less than 24.5 N, so yes.
Kinetic friction. f_k = 0.30·42.44 = 12.73 N up the slope.
Acceleration. a = (24.5 − 12.73)/5.0 = 2.35 m/s² down the slope.
Check. Smaller than the frictionless value g sin 30° = 4.9 m/s². ✓

Common mistakes

Writing N = mg on a ramp. Here N = mg cos θ.
Always using f = μN. Static friction is only as big as it needs to be.
Drawing "ma" as a force on the free-body diagram.

A recipe for any force problem

Isolate one object. Draw only the forces acting on it: gravity, contact forces (normal and friction), tensions, pushes.
Pick axes that fit the motion. On a ramp, tilt them so one axis runs along the slope. Then the acceleration has only one component.
Write ΣF = ma for each axis. Perpendicular to the ramp nothing accelerates, so N = mg cos θ here. N is not always mg.
Third-law pairs act on different objects. The normal force on the block and gravity on the block are not a pair. The pair of "Earth pulls block" is "block pulls Earth."

Check yourself: a book rests on a table. The third-law partner of "Earth's gravity pulls the book down" is:

Week

Which way does friction point?

The most common free-body-diagram mistake is drawing friction the wrong way. One test settles it every time: on ramps, under wheels, on rolling balls and under ladders.

The intuition: the ice test

Friction is the surface's way of resisting slipping. So imagine the surface turned to ice. Ask: which way would the contact point slide? Friction points the other way.

Friction doesn't oppose "the motion". It opposes slipping between the two surfaces. Usually those match, but not always: the road pushes a speeding-up car forward, and a ball rolling up a ramp feels friction pointing up.

Question	If yes	If no
Is the contact point actually sliding over the surface?	Kinetic friction: size μₖN, pointing opposite the sliding	Static friction: any size up to μₛN, whatever is needed
Static: would it slide on ice? (no friction at all)	Friction points opposite that would-be slide	No friction needed (for example, steady rolling on level ground)
Rolling without slipping? (v = ωR)	The contact point is at rest, so static. It does no work.	Slipping: compare v with ωR to see which way the bottom slides, then kinetic

Wheel explorer: rolling, skidding or spinning out?

—

Forward speed v

Spin speed at the rim ωR

If it rolls without slipping, what is applied?

Contact point speed (v − ωR)—

Friction type—

Friction points—

Predict the direction: 14 cases

Look at the picture, use the ice test, then choose. The free-body diagram updates with friction in place.

1. A box sits still on a ramp. Nothing pushes it.

2. You push a box up a ramp, hard, but it hasn't started moving.

3. A box was shoved and is now sliding UP the ramp, slowing down.

4. A box is sliding DOWN the ramp.

5. A box rides on the bed of a truck that is speeding up. The box doesn't slip.

6. Two stacked blocks. You pull the bottom one to the right and the top one moves along without slipping. Which way is friction on the TOP block?

7. A ball rolls DOWN a ramp without slipping.

8. A ball rolls UP a ramp without slipping, slowing down.

9. A car speeds up from rest on a flat road. The tires grip.

10. A car brakes so hard the wheels lock and it skids.

11. A bowling ball is released sliding forward with no spin.

12. A ball rolls at constant speed across a level floor.

13. A ladder leans against a frictionless wall. Which way does the floor's friction push on the ladder's foot?

14. A coin rides on a turntable spinning at a steady rate. The coin doesn't slip.

Try this

Predict what will happen first. Then set it up and check.

Make the wheel spin exactly as fast as it moves.
What you should see
The contact point is at rest: rolling without slipping, and with nothing applied, no friction at all.
Spin the wheel much faster than it moves (a burnout).
What you should see
The contact point slides backward, so kinetic friction points forward. It pushes the car ahead and slows the spin.
No spin at all, like a fresh bowling ball.
What you should see
The contact point slides forward, so kinetic friction points backward.
Roll without slipping, then add engine torque.
What you should see
Static friction appears and points forward. It is the road pushing the car ahead.

Worked example

A 5.0 kg box sits on a 35° ramp with μₛ = 0.30. You push up along the ramp. (a) What is the smallest push that keeps it from sliding down? (b) What is the largest push before it starts sliding up?

Try it yourself first, then reveal one step at a time.

(a) Small push: the box wants to slide down, so friction helps hold it up the slope.

(b) Big push: the box wants to slide up, so friction points down the slope.

Picture and FBDs. Forces: mg, N, your push P (up the slope) and static friction. Its direction depends on which way the box would slide on ice, so draw two cases.
Numbers along and across the ramp. mg sin 35° = 49·0.574 = 28.1 N down the slope. N = mg cos 35° = 40.1 N, so the most static friction can give is μ_sN = 12.0 N.
(a) About to slide down. Friction points up and is at its maximum: P + 12.0 = 28.1, so P_min = 16.1 N.
(b) About to slide up. Friction points down and is at its maximum: P = 28.1 + 12.0 = 40.1 N.
Interpret. For any push between 16.1 N and 40.1 N the box stays put, and static friction quietly adjusts its size and even its direction: up the slope below 28.1 N, down the slope above it.
Check. At exactly P = 28.1 N, friction is zero. ✓

Common mistakes

"Friction always points opposite the motion." True for kinetic friction. Static friction opposes the tendency to slip, which can be along the motion (a car speeding up, a ball rolling up a ramp).
Using μₛN for static friction automatically. It is only that big on the verge of slipping.
Giving rolling objects kinetic friction. If there is no slipping, the contact point is at rest, so it is static.
Forgetting the 3rd-law partner: if the floor pushes the ladder toward the wall, the ladder pushes the floor away from it.

The rules in one place

Kinetic: f_k = μ_kN, opposite the sliding of the contact point relative to the surface.
Static: f_s ≤ μ_sN, direction from the ice test, size from ΣF = ma (and Στ = Iα when things spin).
Ramps: holding a box still with no push, friction points up the slope. Push hard enough and it flips to point down.
Rolling: rolling down or up a ramp, static friction points up the slope. On level ground at constant speed, none. Driven wheels speeding up: forward. Braking without skidding: backward.
Slipping: compare v with ωR. If v > ωR the bottom slides forward (friction back). If ωR > v it slides backward (friction forward).
Extended objects (ladders, rods): find the direction with the ice test at each contact, then get the size from a torque balance.

Check yourself: a car is speeding up on a dry road. Friction from the road on the driven tires points:

Week

Strings and pulleys

When objects are tied together they speed up together. Treat them as one team to find the acceleration, then look at one member to find the pull in the string.

The intuition

Picture a train: the engine pulls, the cars follow, and every car has the same speed. A tight string turns connected objects into a train. Here the hanging weight is the engine, friction on the table is a brake, and the total mass is everything that has to be moved.

The string pulls on both objects with the same tension. On the hanging mass it pulls up, which is why that mass falls more slowly than if you just dropped it.

Simplification for now

We assume: the string is massless and doesn't stretch; the pulley is massless and frictionless
Which lets us: use one tension T on both sides of the pulley, and give both objects the same acceleration
This changes in: Week 6, Rolling (Pulleys revisited): a pulley with mass needs a net torque to spin up, so T₁ ≠ T₂

Block on a table, pulled by a hanging mass

—

Block on table m₁

Hanging mass m₂

Friction μ

Acceleration—

Tension T—

m₂g—

a = (m₂ − μm₁)g / (m₁ + m₂), T = m₂(g − a)

Notice T is less than m₂g while the hanging mass speeds up. If the string pulled up with the full m₂g, the hanging mass wouldn't accelerate.

Try this

Predict what will happen first. Then set it up and check.

No friction, equal masses.
What you should see
a = g/2, and the tension is half of m₂g.
Make the hanging mass much heavier.
What you should see
The acceleration gets close to g. The light block barely holds it back.
Add lots of friction.
What you should see
Nothing moves once μm₁g is at least m₂g.

Worked example

A 4.0 kg block on a table (μₖ = 0.25) is tied over a light pulley to a hanging 2.0 kg mass. Find the acceleration and the tension.

Try it yourself first, then reveal one step at a time.

Block on the table.

Hanging mass.

Picture and FBDs. Block: tension T forward, friction f back, N up, m₁g down. Hanging mass: T up, m₂g down.
Whole system. Driving force m₂g = 19.6 N, resisting force μₖm₁g = 9.8 N, total mass 6.0 kg. a = (19.6 − 9.8)/6.0 = 1.63 m/s².
Hanging mass alone. m₂g − T = m₂a, so T = 2.0(9.8 − 1.63) = 16.3 N.
Check with the block. T − f = 16.3 − 9.8 = 6.5 N and m₁a = 4.0·1.63 = 6.5 N. ✓
Sense check. T is less than m₂g = 19.6 N, as it must be while the hanging mass speeds up. ✓

Common mistakes

Setting the tension equal to m₂g while the hanging mass accelerates.
Forgetting that the string pulls on both objects with the same tension.
Using only the hanging mass in the denominator. The whole team moves together.

A two-step method that always works

Step 1, the whole team: (forces that drive it − forces that resist it) ÷ total mass. Here a = (m₂g − μm₁g)/(m₁ + m₂).
Step 2, one object alone: draw the hanging mass by itself: m₂g − T = m₂a. Solve for T.
Ideal string and pulley: one tension along the whole string, and both ends move the same distance.
Constraints: if a rope runs around a movable pulley, the pulley moves half as far as the free end of the rope, so its acceleration is half as big. Write that relation down before solving.

Check yourself: while the hanging mass speeds up on its way down, the string tension is:

WeekCalculus

Work and energy

Energy is bookkeeping. When you don't care about time, tracking energy is usually much faster than tracking forces.

The intuition

Picture a marble rolling on a curved track with no friction. The track's height is the potential energy U(x). The marble's total energy E is a horizontal line it can never rise above. The gap between E and the track is kinetic energy: big gap, fast marble.

Where the track meets the line, the marble stops and turns around. Where the track slopes downhill, the force pushes downhill. That is F = −dU/dx: force is the downhill slope of energy.

Assumption upgrade

Before: forces moment by moment, with friction only on flat surfaces
So we didn't need: to track energy
Now: paths curve and forces change direction, which makes F = ma hard
So we need: energy bookkeeping: K + U is constant without friction, and drops by f_kd with it

W = ∫ F · drwork

W_net = ΔKwork–energy

F_x = −dU/dxforce from U

P = F · vpower

Total energy E

—

x—

Kinetic K—

Potential U—

Total E—

Force −dU/dx—

Green shading marks where the particle is allowed (E > U); its height is K. Raise E in the double well above the central hump and the particle starts visiting both sides.

Try this

Predict what will happen first. Then set it up and check.

Double well: raise E above the hump.
What you should see
Below 4 J the particle stays in one valley; above it, it crosses between both.
Watch where the particle is fastest.
What you should see
At the bottom of the valley, where the gap between E and U (the kinetic energy) is biggest.
Molecular bond: raise E above 0.
What you should see
The atom escapes and never comes back. The bond breaks.

Worked example

A 0.50 kg block slides from rest down a frictionless curved ramp 3.0 m high, then across a rough floor with μₖ = 0.20. How far does it slide on the floor?

Try it yourself first, then reveal one step at a time.

On the rough floor.

Picture and FBD. On the ramp the normal force is perpendicular to the motion, so it does no work. On the floor, friction does negative work.
Energy at the bottom of the ramp. mgh = ½mv², so v = √(2·9.8·3.0) = 7.67 m/s.
Friction takes it all. On the floor N = mg, so f_kd = μ_kmgd must equal mgh.
Solve. d = h/μ_k = 3.0/0.20 = 15 m.
Check. The mass and the ramp shape both cancel. Units: meters. ✓

Common mistakes

Using energy to find a time. Energy gives speeds and heights.
Getting the sign of F = −dU/dx backward. Force points downhill on the U graph.
Forgetting energy lost to friction.

Reading energy diagrams and using energy

Turning points are where E = U. Equilibria are where the slope is zero: valleys are stable, hilltops unstable.
Conservative forces (gravity, springs) store energy you get back: U_g = mgh, U_s = ½kx². Friction doesn't; it turns mechanical energy into heat, so K + U drops by f_kd.
Variable forces: work is the area under F versus x. A spring stretched to x has stored ∫₀^x kx′ dx′ = ½kx².
When to use energy: the question asks about speed or height, not time. Use forces or kinematics when you need time or direction.

Check yourself: at a point where U(x) slopes upward to the right, the force on the particle points:

Week

Energy in action

Energy tells you how fast something is going. Forces tell you whether the track can hold on. Many problems need both.

The intuition

Think of a roller coaster: the higher it starts, the faster it is at the bottom, because height trades for speed. At the top of a loop, though, the cart is upside down. It only stays on the track if it is moving fast enough that the track still has to push on it.

Release it too low and it slows near the top, the track stops pushing, and the cart peels away and falls.

Simplification for now

We assume: the cart slides without friction and does not roll
Which lets us: use ½mv² as its only kinetic energy, giving h ≥ 2.5R
This changes in: Week 6, Rolling: a rolling solid ball also stores spin energy and needs h ≥ 2.7R

Loop-the-loop: how high must you start?

—

Release height h, in loop radii

Speed at top—

Normal force at top—

Needed h2.5 R

Energy gives the speed anywhere: v² = 2g(h − y). Forces decide whether the track can hold you: at the top the track can only push down, so you need v² ≥ gR. Combining the two gives h ≥ 2.5R, not 2R. Energy alone is not enough here.

Try this

Predict what will happen first. Then set it up and check.

Release from exactly 2R.
What you should see
It falls off before the top, about 132° around.
Release from exactly 2.5R.
What you should see
It just makes it: at the top the track pushes with zero force.
Release from below R.
What you should see
It never gets above the middle of the loop and slides back down.

Worked example

A 50 kg cart slides without friction from height h into a vertical loop of radius R = 2.0 m. (a) What is the smallest h that works? (b) If h = 6.0 m, what normal force acts at the top?

Try it yourself first, then reveal one step at a time.

At the top of the loop (upside down): both forces point down, toward the center.

Picture and FBD at the top. Gravity and the track's normal force both point down, toward the center: N + mg = mv²/R.
Condition to stay on. N ≥ 0, so v² ≥ gR at the top.
Energy. From release to the top (height 2R): v² = 2g(h − 2R).
(a) Combine. 2g(h − 2R) ≥ gR gives h ≥ 2.5R = 5.0 m.
(b) h = 6.0 m. v² = 2·9.8·2.0 = 39.2 m²/s², N = 50(39.2/2.0 − 9.8) = 490 N.
Check. At exactly h = 5.0 m, N would be 0. Higher start, positive N. ✓

Common mistakes

Thinking h = 2R is enough because the top is at 2R.
Forgetting that gravity and the normal force both point down at the top.
Using kinematics equations on a curved track. Use energy.

Energy problems, step by step

Roller coasters: with no friction, the speed at height y is v² = 2g(h − y). The shape of the track in between doesn't matter.
Spring launchers: spring energy turns into speed or height: ½kx² = ½mv² or = mgh.
The loop: at the top both gravity and the track push down, so N + mg = mv²/R. Staying on needs N ≥ 0, so v² ≥ gR. Energy then gives h ≥ 2.5R.
With friction: subtract the lost energy, f_kd, from the total.

Check yourself: a cart is released from height 2R into a frictionless loop of radius R. What happens?

Week

Center of mass

Every system, however messy, has one point that moves as if all the mass were there and all the outside forces pushed on it. Knowing when to use it turns hard problems into short ones.

The intuition

Balance a ruler on one finger: the balance point is its center of mass. For a group of objects it is the mass-weighted average position, so heavy things pull it toward themselves.

The big idea: forces between parts of a system can't move its center of mass. You can't lift yourself by your shoelaces, and walking on a floating canoe can't move the person + canoe center. Only outside forces can.

Assumption upgrade

Before: each problem had one object, treated as a point
So we didn't need: to think about where in a system the mass is
Now: systems of many objects and extended bodies
So we need: the center of mass, and the rule that only external forces change its motion

x_cm = Σm_ix_i / Σm_idiscrete

x_cm = (1/M)∫x dmcontinuous

ΣF_ext = Ma_cmsystem's 2nd law

P_total = Mv_cmmomentum

Should I use the center of mass?

Look for these clues in the problem statement.

You see…	Why the center of mass helps	What to write
People walking on a boat, raft or cart; things pushing apart from rest on ice or in space	No outside horizontal force, so the center of mass can't move	x_{cm, before} = x_{cm, after}
An explosion, a firework, a rocket stage, a projectile that splits	The explosion forces are internal, so the center of mass keeps its old motion	Follow the CM along the original path, then solve for the missing piece
A plank, ladder, sign, rod or body in a torque or balance problem	Gravity acts as if all the weight sits at the CM	Put Mg at x_cm in Στ = 0
A chain, rope, rod or person rising or falling	Gravitational energy depends only on the CM height	ΔU = MgΔy_cm
"The whole system": blocks pushing each other, a train of carts	Internal forces cancel in pairs	ΣF_ext = Ma_cm, one FBD for everything
Collisions, especially elastic ones	Total momentum is Mv_cm, and it never changes	Switch to the CM frame: elastic velocities just reverse
Rolling or tumbling objects	Motion splits into CM motion plus spin about the CM	K = ½Mv_cm² + ½I_cmω²
One small object (a ball, a car treated as a point)	It already is its center of mass	No CM step needed: use ΣF = ma or energy directly

Canoe walker: still water, no friction from it sideways

Person's spot on the canoe

Person's mass m

Canoe mass M

Canoe moved—

Person moved (vs water)—

Center of mass moved—

Exploding shell: it bursts at the top of its arc

Time

Front piece share of the mass

Explosion strength (separation speed)

Front piece lands at—

Back piece lands at—

CM lands at—

No-explosion range—

Which tool? Spot the center-of-mass problems

For each situation, pick the idea that cracks it.

A 60 kg person walks from one end of a 120 kg canoe to the other on still water. How far does the canoe move?

A firework shell explodes at the top of its arc into three pieces. Where does the center of the pieces land?

A ladder leans against a frictionless wall. How much friction does the floor need to supply?

You push two boxes in contact across a floor with force F. What is their acceleration?

A thrown ball: how high does it go?

An astronaut floating at rest throws a wrench. Where is the astronaut + wrench center of mass a minute later?

A uniform chain hangs halfway off a table. How much work does it take to pull it all back up?

A rod pivoted at one end swings down from horizontal. How fast does it spin at the bottom?

A rocket coasting at 200 m/s in deep space separates into two stages with a spring. How fast does their combined CM move afterward?

Two carts are tied together and pulled with force F. What is the tension in the rope between them?

Try this

Predict what will happen first. Then set it up and check.

Walk the person all the way to the far end.
What you should see
The canoe slides back 1.33 m while the person moves 2.67 m forward. The triangle (CM) never moves.
Make the canoe very heavy.
What you should see
It barely moves. A heavy canoe needs only a small shift to balance the person's move.
Make the person and canoe equally heavy.
What you should see
Each moves 2 m: equal masses split the distance equally.
Exploding shell: the classic problem.
What you should see
The back piece drops straight down; the front piece lands 1.5 times as far as the shell would have. The CM still lands at the original spot.
Make the front piece tiny.
What you should see
The tiny piece flies far, the big one barely changes course. The CM still lands at the same spot.

Worked example

A 60 kg person stands at the left end of a 4.0 m, 120 kg canoe at rest on still water. They walk to the right end. How far does the canoe move, and which way?

Try it yourself first, then reveal one step at a time.

Person + canoe as one system: only vertical outside forces.

Picture and FBD. Treat the person and canoe as one system. Outside forces: gravity down and the water's buoyancy up. No horizontal outside force (ignore water drag), and it starts at rest, so x_cm cannot change.
Before. Origin at the canoe's starting left end. Person at 0, canoe's center at 2.0 m: x_cm = (60·0 + 120·2.0)/180 = 1.33 m.
After. The canoe has shifted by D. Person at D + 4.0, canoe center at D + 2.0: x_cm = (60(D + 4.0) + 120(D + 2.0))/180 = D + 2.67 m.
Set them equal. D + 2.67 = 1.33, so D = −1.33 m: the canoe moves 1.33 m to the left.
Person's move. Relative to the water the person moved 4.0 − 1.33 = 2.67 m right.
Check. Mass × displacement balances: 60·2.67 = 120·1.33 = 160 kg·m. ✓

Common mistakes

Using the canoe's length as how far the person moves relative to the water. Both move, in opposite directions.
Thinking an explosion changes where the center of mass lands. Only outside forces can do that.
Putting a plank's weight at its end instead of its middle.

Finding the center of mass

Symmetry first: a uniform rod, disk, sphere or box has its CM at its geometric center.
Composite objects: replace each piece by a point mass at its own CM, then average. A hole counts as negative mass.
Continuous objects: x_cm = (1/M)∫x dm with dm = λ dx. For a rod whose density grows along it, the CM shifts toward the heavy end.
The CM doesn't have to be inside the object: a ring's or a boomerang's is in empty space.
Linked ideas: the CM frame for collisions, the parallel axis theorem I = I_cm + Md² for rolling, and weight at the CM for torque.

Check yourself: you walk toward the front of a floating canoe. What does the canoe do?

Week

Momentum and collisions

In a collision the forces are huge, brief and messy. Momentum lets you skip them entirely: whatever one object gains, the other loses.

The intuition

Momentum is "how hard is this to stop?" A truck at walking speed and a baseball at 90 mph can both hurt. In a collision the two objects push on each other equally and for the same time, so the momentum one loses the other gains. The total stays fixed.

Energy is pickier. Bouncy collisions keep kinetic energy; squishy ones turn some into heat and dents. Through all of it, the center of mass glides on at a steady speed, as if nothing happened.

p = mvmomentum

J = ∫F dt = Δpimpulse

m₁v₁ + m₂v₂ = m₁v₁′ + m₂v₂′conserved

View from the CM frame

Time, collision at 0

In 3D The same collision on a track. Spin the view to see both carts and the center of mass (the black marker under the track). The CM keeps moving at the same velocity before and after the hit, because the collision forces are internal.

Cart 1 mass

Cart 1 velocity

Cart 2 mass

Cart 2 velocity

Bounciness e

v₁ after—

v₂ after—

Total p—

K before—

K after—

K lost—

—

The triangle under the track is the center of mass. It moves at the same speed before and after, whatever e is. Try equal masses with e = 1: the carts swap velocities. Now tick "CM frame": riding along with the center of mass, total momentum is zero, and an elastic collision just reverses each cart's velocity. That is the easy way to solve elastic collisions.

Try this

Predict what will happen first. Then set it up and check.

Equal masses, perfectly bouncy.
What you should see
The carts swap velocities.
Make them stick together.
What you should see
Kinetic energy is lost but the center of mass keeps exactly the same speed.
Watch from the center-of-mass frame.
What you should see
Each cart just reverses its velocity. Total momentum in this frame is zero.

Worked example

A 3.0 kg cart moving at 4.0 m/s hits a 1.0 kg cart at rest and they stick together. Find their final speed and the kinetic energy lost.

Try it yourself first, then reveal one step at a time.

Cart 1 during the collision.

Cart 2: equal and opposite push (3rd law).

Picture and FBDs. During the hit the carts push on each other with equal and opposite forces. Vertical forces cancel. No outside horizontal force, so total momentum is conserved.
Momentum. 3.0·4.0 + 1.0·0 = (3.0 + 1.0)v, so v = 3.0 m/s.
Kinetic energy before. ½·3.0·4.0² = 24 J.
After. ½·4.0·3.0² = 18 J. Lost: 6.0 J (25%), turned into heat and deformation.
Check. The final speed is between 0 and 4 m/s, and KE decreased, as it must when objects stick. ✓

Common mistakes

Assuming kinetic energy is conserved in every collision.
Dropping signs. A cart moving left has negative momentum.
Conserving momentum while an outside force acts for a long time.

Using momentum

When is it conserved? When the net external impulse is negligible. During a short collision, gravity and friction barely matter, so momentum is conserved across the collision even if not before or after.
Elastic (e = 1): kinetic energy also conserved. Perfectly inelastic (e = 0): they stick, and the most kinetic energy possible is lost. Momentum is conserved in both.
2D collisions: conserve each component separately.
Impulse is the area under F versus t. Airbags and bent knees work by stretching the time, which lowers the force for the same Δp.
Center of mass (full module: when to use it): x_cm = Σmx / Σm, and ΣF_ext = Ma_cm.

Check yourself: two carts collide and stick together. Which is conserved?

Week

Impulse and rockets

Impulse is a push added up over time. It explains why airbags and bent knees protect you, and how a rocket speeds up with nothing to push against.

The intuition

Stopping an egg always takes the same change in momentum. You only choose how: a hard floor stops it in a few milliseconds with a huge force; a hand that gives stretches the stop over a tenth of a second with a small force. The area under the force-versus-time graph is the same either way.

A rocket is the same idea run backward: it throws exhaust out the back, and the exhaust pushes back on the rocket.

Assumption upgrade

Before: masses never changed, so F = ma worked
So we didn't need: to think about mass entering or leaving
Now: a rocket throws away mass as it goes
So we need: the general form F = dp/dt, applied to a system that includes the exhaust

Stopping time Δt

Mass

Speed before the stop

Impulse = Δp—

Average force—

Peak force—

Average ÷ weight—

The rocket equation: momentum with changing mass

The intuition

A rocket throws mass backward to push itself forward. Early on, each kilogram of exhaust has to shove the whole fuel load along with it, so it buys very little speed. Late in the burn the rocket is light and the same kilogram buys much more. Adding up those diminishing returns gives a logarithm.

Exhaust speed u

Mass ratio m₀/m_f

Speed gained Δv—

Launch mass that is fuel—

Ratio for 7.8 km/s—

dp/dt = F with changing mass ⟹ m dv = −u dm ⟹ Δv = u ln(m₀/m_f)

Try this

Predict what will happen first. Then set it up and check.

Drop the egg on a hard floor.
What you should see
A few milliseconds, a force more than a hundred times the egg's weight.
Now catch it in a pillow.
What you should see
Same area under the curve, but the force is a tiny fraction of before.
Rocket: set the mass ratio to 10 and note Δv, then double it to 20.
What you should see
Δv goes from u ln 10 ≈ 2.3u to u ln 20 ≈ 3.0u: doubling the mass ratio adds only u ln 2 ≈ 0.7u. Piling on fuel helps much less than you'd expect.

Worked example

A 0.15 kg baseball arrives at 40 m/s and leaves the bat at 50 m/s in the opposite direction. The contact lasts 1.5 ms. Find the average force from the bat.

Try it yourself first, then reveal one step at a time.

Ball during contact. The bat force is enormous; gravity is tiny.

Picture and FBD. During contact the bat pushes the ball back toward the pitcher. Gravity (1.5 N) is negligible next to it.
Pick a direction. Toward the pitcher is +. Before: −40 m/s. After: +50 m/s.
Impulse. Δp = 0.15(50 − (−40)) = 13.5 kg·m/s.
Average force. F = Δp/Δt = 13.5/0.0015 = 9000 N, about 6000 times the ball's weight.
Check. Just stopping the ball would need only 6 kg·m/s; the bounce more than doubles the impulse. ✓

Common mistakes

Thinking a soft landing reduces the change in momentum. It reduces the force.
Forgetting that a bounce doubles the impulse.
Using F = ma with constant mass for a rocket.

Using impulse

J = ∫F dt = Δp. The average force during a collision is Δp/Δt.
Bounces hurt more: reversing direction doubles Δp compared with stopping, so the impulse (and the force, for the same time) doubles.
Variable mass (rockets, a cart catching rain): use F = dp/dt on a system that includes the mass coming or going. A cart with rain falling into it must be pushed with F = v dm/dt just to keep its speed.
Rocket equation: Δv = u ln(m₀/m_f). Doubling the fuel does not double the speed gain.

Check yourself: why do airbags reduce injuries?

Week

Torque and balance

A torque is a force's ability to make something turn. It depends on how hard you push and how far from the pivot you push.

The intuition

On a seesaw, a small kid far from the middle can balance a big kid sitting close in. What matters for turning is weight × distance from the pivot. That product is the torque.

It is the same reason a long wrench loosens a stuck bolt, and why door handles are on the side away from the hinges.

Assumption upgrade

Before: objects were points, so every force acted at the same place and only ΣF mattered
So we didn't need: to know where a force acts
Now: objects have size
So we need: torque (where and at what angle a force acts), Στ = 0 for balance, and the weight placed at the center of mass

τ = rF sin θtorque

Στ = Iαrotational 2nd law

m₁d₁ = m₂d₂seesaw balance

Seesaw: drag the blocks or use the sliders

—

In 3D Torque is a vector along the axle. Each weight's torque r × F points along the pivot axis, the two in opposite directions; the bigger one wins, and the plank turns the way the fingers of your right hand curl when your thumb points along the net torque.

Left mass m₁

Left distance d₁

Right mass m₂

Right distance d₂

Left torque (CCW)—

Right torque (CW)—

Net torque—

Angular accel. at release (incl. 2 kg, 6 m beam)—

Try this

Predict what will happen first. Then set it up and check.

Balance the seesaw by moving the light block.
What you should see
At d₂ = 3 m: 3 kg × 1 m = 1 kg × 3 m.
Put equal masses at unequal distances.
What you should see
The far one wins. Distance matters as much as weight.
Move both masses close to the pivot.
What you should see
Even unbalanced, the angular acceleration is bigger: less moment of inertia to turn.

Worked example

A 4.0 m, 10 kg uniform plank rests on supports at both ends. A 60 kg person stands 1.0 m from the left end. Find the force from each support.

Try it yourself first, then reveal one step at a time.

The plank: two support forces up, two weights down.

Picture and FBD. Forces on the plank: N_L and N_R up at the ends, its own weight (98 N) at the middle, the person's weight (588 N) 1.0 m from the left.
Forces balance. N_L + N_R = 98 + 588 = 686 N.
Torques about the left end. N_L has no lever arm there, so it drops out: N_R·4.0 = 98·2.0 + 588·1.0 = 784 N·m.
Solve. N_R = 196 N, N_L = 686 − 196 = 490 N.
Check. The person stands nearer the left end, so the left support carries more. ✓

Common mistakes

Using the full length of the beam instead of the distance from the pivot.
Forgetting the sin θ: only the perpendicular part of a force makes torque.
Mixing up clockwise and counterclockwise signs.

Working with torque

Lever arm: only the part of the force perpendicular to the arm turns things. Push straight along a door toward its hinge and nothing happens (sin θ = 0).
Signs: pick counterclockwise as positive (or clockwise) and stick with it.
Static equilibrium needs both ΣF = 0 and Στ = 0. The torque balance works about any point, so choose one where an unknown force acts.
Angular acceleration: α = Στ/I. Mass far from the pivot makes I bigger and the response slower.

Check yourself: a 30 kg child sits 1.0 m from the pivot of a seesaw. Where must a 20 kg child sit to balance?

WeekCalculus

Rotation and rolling

Every idea from straight-line motion has a rotational twin. Learn the dictionary and you already know most of rotation.

The intuition

Moment of inertia is mass, but where the mass sits matters. Mass far from the axis is harder to spin up, the way a door is hard to push near its hinge. A hoop puts all its mass at the rim; a solid ball keeps most of it near the middle.

When something rolls downhill, gravity's energy has to pay for two things: moving forward and spinning. A hoop spends half on spinning, so it moves forward slowest. Mass and size cancel; only the shape matters.

Assumption upgrades in this module

Before we assumed	So we didn't need	Now	So we need
pulleys were massless (Week 2), so the tension was the same on both sides	a torque balance on the pulley	pulleys have mass and must be spun up	T₁ ≠ T₂, with (T₂ − T₁)R = Iα. Try Pulleys revisited below
objects slid, so kinetic energy was only ½mv²	rotational energy, or friction that makes things spin	objects roll	K = ½mv² + ½Iω², the rolling condition v = ωR, and static friction at the contact point
friction always opposed the motion (kinetic friction, or static friction holding a block still)	to think about which way friction points on a spinning object	for rolling without slipping, the contact point is momentarily at rest	static friction, which can point up or down a ramp, does no work, and is only as big as needed. See Which way does friction point?
the loop-the-loop cart slid (Week 4), needing h ≥ 2.5R	spin energy	a solid ball rolls through the loop	h ≥ 2.7R, because 2/7 of its kinetic energy is spin

Straight line	Rotation	Link
x, v, a	θ, ω, α	v = ωr, a_t = αr
mass m	moment of inertia I = Σmr²	I = I_cm + Md² (parallel axis)
force F	torque τ = rF sin φ	τ = r × F
F = ma	τ = Iα	same structure
½mv²	½Iω²	rolling: both at once

Race down a ramp, released from rest

Ramp angle θ

Ramp length

Object	I / MR²	a (m/s²)	Time (s)	Speed at bottom	KE that spins

Rolling without slipping: a = g sin θ / (1 + β), β = I/(MR²)

Pulleys revisited: give the pulley some mass

—

Left mass m₁

Right mass m₂

Pulley mass M (uniform disk)

Acceleration—

T₁ (left side)—

T₂ (right side)—

T₂ − T₁—

Why can the tensions differ? The rope doesn't slip on the pulley, so static friction between rope and rim grips the pulley. The difference in tension is the torque that spins it up: (T₂ − T₁)R = Iα = ½MR²(a/R), so T₂ − T₁ = ½Ma.

Try this

Predict what will happen first. Then set it up and check.

Press Race.
What you should see
The block slides fastest; among the rollers, the solid sphere wins and the hoop comes last.
Steepen the ramp and race again.
What you should see
Faster race, same finishing order. Only shape matters.
Read the "KE that spins" column.
What you should see
The hoop puts 50% of its energy into spinning, the solid sphere only about 29%.

Worked example

A solid cylinder (2.0 kg, I = ½MR²) rolls without slipping from rest down a 30° ramp. Find its acceleration and the friction force.

Try it yourself first, then reveal one step at a time.

Cylinder on the ramp. Static friction acts at the contact point, up the slope.

Picture and FBD. Forces: mg down, N perpendicular to the ramp, static friction up the slope at the contact point. Friction is what makes it spin.
Translation. mg sin θ − f = ma.
Rotation about the center. fR = Iα = ½MR²(a/R), so f = ½ma.
Combine. mg sin θ = (3/2)ma, so a = (2/3)(9.8)(0.5) = 3.27 m/s².
Friction. f = ½·2.0·3.27 = 3.27 N.
Check. Less than g sin 30° = 4.9 m/s² (a sliding block's value). ✓

Common mistakes

Forgetting rotational kinetic energy for a rolling object.
Using v = ωR when the object slips.
Using I about the wrong axis. Shift it with the parallel axis theorem.

Where the formula comes from

Along the ramp: Mg sin θ − f = Ma. About the center: fR = Iα = Ia/R, using the rolling condition a = αR. Eliminate f and M drops out. Static friction makes it roll but does no work, because the contact point is momentarily at rest.

The moment of inertia comes from an integral, I = ∫r² dm. For a thin rod about its center, I = ∫_−L/2^L/2 x² (M/L) dx = ML²/12. About its end it is ML²/3, which the parallel axis theorem confirms.

Check yourself: a heavy hoop and a light solid sphere roll down the same ramp. Which reaches the bottom first?

Week

Angular momentum

Spinning things keep spinning, and they keep their angular momentum even when they change shape. That one rule explains figure skaters, diving and why the Earth's day barely changes.

The intuition

Angular momentum is "how much spin" an object carries. With no outside twist, that amount can't change. Pull mass inward and it has to spin faster to keep the same total, like a skater pulling in their arms.

Spinning faster means more kinetic energy. That energy comes from you: pulling the masses inward against their tendency to fly outward takes work.

Assumption upgrade

Before: momentum was p = mv for point particles
So we didn't need: spin
Now: objects spin and change shape
So we need: L = Iω (or r × p), conserved when there is no external torque, not just no external force

L = Iωrigid body

L = r × ppoint particle

τ_net = dL/dtrotational 2nd law

Spinning platform with two masses, top view

In 3D The angular momentum L = Iω points up the axle (right-hand rule). With no outside torque its length never changes: move the masses in and ω goes up, move them out and ω goes down.

Pull masses in or out, r

Each mass m

Platform I₀

I = I₀ + 2mr²—

ω—

L (constant)—

Kinetic energy—

Changing m or I₀ starts a new spin at 2 rad/s. Moving r keeps L fixed: watch ω and the kinetic energy rise as the masses come in.

Right-hand rule and precession, in 3D

Which way does a torque point?

Point your fingers along r, from the pivot (the bolt) to where the force pushes.
Curl them toward F, the short way around.
Your thumb points along τ = r × F. The wrench turns the way your fingers curl.

Angle from r to F

τ = rF sin θ—

τ points—

Here r = 0.50 m and F = 20 N. Try 180°: pushing straight along the wrench gives no torque. Past 180° your thumb flips down, and the wrench turns the other way.

Try this

Predict what will happen first. Then set it up and check.

Pull the masses in.
What you should see
ω climbs, L stays the same, and the kinetic energy goes up.
Push them back out.
What you should see
The kinetic energy drops back. The energy you put in by pulling comes back out.
Make the platform heavy, then pull in.
What you should see
A smaller effect: the movable masses are now a smaller share of the total I.

Worked example

A merry-go-round (uniform disk, 100 kg, radius 2.0 m) spins at 1.0 rad/s. A 50 kg child, initially not moving, jumps onto its edge. What is the new angular speed?

Try it yourself first, then reveal one step at a time.

Merry-go-round plus child. Both forces act along the axle.

Picture and FBD. External forces on the system are gravity and the axle's push, both along the axle. Neither exerts a torque about it, so angular momentum is conserved.
Moments of inertia. Disk: ½MR² = 200 kg·m². Child at the edge: mR² = 200 kg·m².
Conserve L. 200·1.0 = (200 + 200)ω, so ω = 0.50 rad/s.
Energy. K goes from 100 J to 50 J. Like a sticky collision, energy is lost even though L is conserved.
Check. Doubling I halved ω. ✓

Common mistakes

Conserving kinetic energy instead of angular momentum.
Conserving linear momentum when a pivot pushes on the system.
Forgetting that L has a direction (right-hand rule).

When angular momentum is conserved

Conserved about an axis when the net external torque about that axis is zero.
Collisions with a pivot (clay sticking to a hinged rod): linear momentum is not conserved, because the pivot pushes. Angular momentum about the pivot is, because the pivot force has no lever arm there.
A particle moving at constant velocity has constant L = r × p about any point, because its perpendicular distance from that point (the lever arm) never changes. A planet's L is constant for a related reason: gravity points at the Sun, so it exerts no torque about the Sun (Week 8).
Kinetic energy can be written K = L²/(2I). Halve I at fixed L and K doubles.
Gyroscopes: a torque perpendicular to L swings L sideways instead of toppling it. The precession rate is Ω = τ/L.

Check yourself: a skater pulls in their arms and their spin rate doubles. Their rotational kinetic energy:

Week

Gravity and orbits

Newton's law of gravity plus F = ma produces all three of Kepler's laws. Launch a satellite sideways and watch them appear.

The intuition

An orbit is falling while missing. Throw a ball sideways and it curves to the ground. Throw it fast enough and the ground curves away as fast as the ball falls. Too slow and the orbit is a squashed ellipse; too fast and it escapes.

Close to the planet, gravity pulls harder and the satellite speeds up. Far away it slows down. Its angular momentum stays fixed, so the line from the planet sweeps out area at a steady rate.

Assumption upgrade

Before: g was a constant 9.8 m/s², so U = mgh
So we didn't need: to know how far from Earth we were
Now: satellites and planets are far away
So we need: F = GMm/r² and U = −GMm/r. Measure r from the center. mgh is only the near-surface approximation

F = GMm/r²gravity

U = −GMm/rzero at ∞

v_circ = √(GM/r), v_esc = √(2GM/r)speeds

T² = (4π²/GM) a³Kepler 3

Satellite launched sideways

Equal-time areas

Launch speed, as a multiple of circular speed

—

In 3D The whole orbit stays in one flat plane. Gravity pulls toward the planet's center, so it exerts no torque about that point: the angular momentum L (perpendicular to the plane) stays constant, and that is why equal times sweep equal areas.

Eccentricity—

Semi-major axis—

Period—

Closest / farthest—

Speed now—

Distances are in units of the launch radius and times in units of the circular-orbit period. Each shaded wedge takes the same time. Narrow wedges near the planet, wide ones far away, same area. For a satellite 400 km up, the circular speed is about 7.7 km/s; escape needs √2 times that.

Try this

Predict what will happen first. Then set it up and check.

Launch at exactly circular speed.
What you should see
A circle. Every wedge has the same shape.
Launch at 60% of it.
What you should see
A narrow ellipse: fast past the planet, slow at the far end.
Launch at 1.45 times it.
What you should see
It escapes. Escape speed is √2 ≈ 1.41 times circular speed.

Worked example

A satellite orbits 400 km above Earth's surface (Earth: R = 6.37 × 10⁶ m, M = 5.97 × 10²⁴ kg). Find its speed and period.

Try it yourself first, then reveal one step at a time.

Satellite: gravity is the only force, pointing at Earth's center.

Picture and FBD. Gravity is the only force. It points at Earth's center and supplies the centripetal force.
Radius from the center. r = 6.37 × 10⁶ + 0.40 × 10⁶ = 6.77 × 10⁶ m.
Newton's second law. GMm/r² = mv²/r, so v = √(GM/r) = 7.67 km/s.
Period. T = 2πr/v = 5550 s ≈ 92 min.
Check. The International Space Station orbits about this high and takes about 92 minutes. ✓

Common mistakes

Measuring r from the surface instead of the center.
Thinking astronauts float because there is no gravity. They are falling along with their spacecraft.
Using U = mgh for orbits.

Kepler's laws, explained by Newton

1. Orbits are ellipses with the planet at one focus. This follows from the 1/r² force; almost any other force law gives orbits that don't close on themselves.
2. Equal areas in equal times. Gravity points straight at the planet, so it exerts no torque and angular momentum is conserved. Area rate = L/(2m).
3. T² ∝ a³. For a circle, set GMm/r² = m(2πr/T)²/r and solve.
Energy: a bound orbit has E = −GMm/(2a) < 0. Escape means E ≥ 0.
Shell theorem: a uniform sphere pulls on outside objects as if all its mass sat at the center. Inside a uniform shell the net pull is zero.

Check yourself: where in an elliptical orbit is the satellite fastest?

Week

Gravity fields

Gravity from several masses just adds up, arrow by arrow. The field is a map of that pull at every point.

The intuition

The field is a map of "which way would a test pebble get pulled, and how hard?" With two masses, each pulls on its own and the pulls add as vectors. Somewhere between them the pulls cancel exactly. That balance point sits closer to the smaller mass.

g = Σ GM_i/r_i² (toward each)field

F = mgforce on a test mass

Gravitational field of two masses (drag them)

Mass ratio M₂/M₁

Zero-field point from M₁—

Earth–Moon (M₂/M₁ = 1/81)90% of the way

Arrow length grows with field strength on a compressed scale. The field g = Σ −GM_ir̂_i/r_i² is the force per kilogram, so F = mg.

Try this

Predict what will happen first. Then set it up and check.

Make the masses equal.
What you should see
The zero-field point sits exactly halfway.
Make M₂ tiny.
What you should see
The balance point moves right up close to M₂.
Drag one mass around.
What you should see
The arrows show the combined pull of both masses, and the zero point stays on the line joining them.

Worked example

Earth and the Moon are 384,000 km apart, and Earth is 81 times as massive. Where between them is the net gravitational field zero?

Try it yourself first, then reveal one step at a time.

A test mass at the balance point: two equal and opposite pulls.

Picture and FBD. A test mass between them feels Earth pulling one way and the Moon the other. At the balance point those pulls are equal.
Set the pulls equal. GM/d₁² = G(M/81)/d₂², with d₁ from Earth and d₂ from the Moon.
Take the square root. d₁/d₂ = 9, and d₁ + d₂ = 384,000 km.
Solve. d₁ = (9/10)·384,000 = 346,000 km from Earth.
Check. Much closer to the lighter Moon, as expected. ✓

Common mistakes

Adding field strengths as plain numbers instead of as arrows.
Using the surface-to-surface distance instead of center-to-center.
Expecting the balance point halfway when the masses differ.

Fields and superposition

Field = force per kilogram. Near Earth's surface it is about 9.8 N/kg downward, which is the same thing as 9.8 m/s².
Superposition: find each mass's pull separately, then add them as vectors.
Shell theorem: a uniform sphere pulls on outside objects as if all its mass sat at the center. Inside a uniform shell, the pulls cancel exactly.
Potential energy also adds: U = −Σ GM_im/r_i, but as plain numbers, not vectors.

Check yourself: where between Earth and the Moon do their gravitational pulls cancel?

WeekCalculus

Simple harmonic motion

Whenever a pull back toward the middle grows in proportion to how far you've strayed, you get a sine wave in time. Springs, pendulums, atoms in a crystal, bridges, all the same equation.

The intuition

Pull a mass on a spring and let go. The spring yanks it back, it overshoots, the spring yanks it the other way. Farther out means a stronger pull, so a big swing covers more ground at higher speed and takes the same time as a small one. The period doesn't depend on amplitude.

Stiffer spring: faster. Heavier mass: slower. So ω = √(k/m) is almost a guess you could make before doing any math.

Simplification for now

We assume: swings are small (θ under about 15°) and there is no friction
Which lets us: replace sin θ by θ, which turns the pendulum into simple harmonic motion with a period that doesn't depend on amplitude
This changes in: Week 10, Damped & driven: friction and air make the oscillation die out

d²x/dt² = −ω²xdefines SHM

x = A cos(ωt + φ)solution

ω = √(k/m), √(g/L)spring, pendulum

In 3D The same oscillator as a real object. Velocity (green) is biggest passing through the middle; acceleration (pink) is biggest at the ends and always points back toward the middle.

Mass m

Spring constant k

Amplitude A

Phase φ

ω—

Period T—

Frequency—

Max speed—

Max accel—

Same color code as Motion graphs. Velocity peaks at the middle, where acceleration is zero; acceleration peaks at the ends, where velocity is zero. The bars show energy sloshing between K and U.

Try this

Predict what will happen first. Then set it up and check.

Double the amplitude.
What you should see
The period stays the same.
Make the mass 4 times bigger.
What you should see
The period doubles, because T ∝ √m.
Pendulum: shorten it to a quarter.
What you should see
The period halves. Mass does not matter for a pendulum.

Worked example

A 0.50 kg block on a spring (k = 50 N/m) on a frictionless table is pulled 0.10 m from equilibrium and released. Find ω, the period, the maximum speed and the maximum acceleration.

Try it yourself first, then reveal one step at a time.

At maximum stretch: the spring pulls back toward equilibrium.

Picture and FBD. Vertical forces cancel. Horizontally only the spring acts: F = −kx, so a = −(k/m)x. That is simple harmonic motion.
Angular frequency. ω = √(k/m) = √100 = 10 rad/s.
Period. T = 2π/ω = 0.63 s.
Maximum speed and acceleration. v_max = Aω = 1.0 m/s, a_max = Aω² = 10 m/s².
Check with energy. ½kA² = 0.25 J and ½mv_max² = 0.25 J. ✓

Common mistakes

Thinking a bigger swing takes longer.
Putting the top speed at the ends. It is in the middle.
Using degrees inside cos(ωt + φ).

Why SHM is everywhere

Near the bottom of any smooth potential-energy valley, U ≈ U₀ + ½U″(x₀)(x − x₀)², which is a spring with k = U″. Small oscillations about any stable equilibrium are simple harmonic.
Pendulum: the restoring torque is −mgL sin θ ≈ −mgLθ for small θ, giving T = 2π√(L/g). Mass cancels.
Physical pendulum: T = 2π√(I/(mgd)), d from pivot to center of mass.
Vertical spring: gravity only shifts the equilibrium; ω is the same.
Energy: E = ½kA², so v_max = Aω and a_max = Aω².

Check yourself: you double the amplitude of a mass on a spring. The period:

WeekCalculus

Damped and driven oscillations

Real oscillators lose energy, and real oscillators get pushed. Together those explain shock absorbers, tuning a radio, and why soldiers break step on bridges.

The intuition

Damping is friction for oscillators: each swing is a fixed fraction smaller than the last, so the amplitude dies away exponentially. Too much damping and it doesn't swing at all; it just oozes back to the middle.

Driving is pushing a kid on a swing. Push at the swing's own rhythm and every push adds energy: the amplitude grows until damping eats energy as fast as you add it. Push at the wrong rhythm and your pushes fight the swing half the time.

Assumption upgrade

Before: oscillators had no friction or air resistance (Week 9)
So we didn't need: energy losses: oscillations went on forever
Now: real oscillators lose energy
So we need: a damping force −bv, which gives decaying amplitude and a finite resonance peak

x″ + γx′ + ω₀²x = (F₀/m) cos ωtγ = b/m

A(ω) = (F₀/m) / √((ω₀² − ω²)² + γ²ω²)steady state

Q = ω₀/γsharpness

Oscillator: time response and resonance curve

Drive on

Natural frequency ω₀

Damping γ

Drive frequency ω

Regime—

Q—

Free ω_d—

Steady amplitude—

Phase lag—

Top: the oscillator starting from rest. Early wiggles are the transient, which dies out at rate γ/2; what remains is the steady response at the drive frequency. Bottom: steady amplitude for every drive frequency; the dot is your setting. Turn the drive off to see free decay and the e^−γt/2 envelope.

Try this

Predict what will happen first. Then set it up and check.

Drive off, lots of damping.
What you should see
It just creeps back with no swing: overdamped.
Drive on, at resonance.
What you should see
The steady amplitude is near its peak and the phase lag is close to 90°.
Now cut the damping way down.
What you should see
The resonance peak gets taller and narrower.

Worked example

A 1.0 kg block on a spring (k = 16 N/m) has a damping force −bv with b = 2.0 kg/s. Is it underdamped? At what frequency does it oscillate, and when has its amplitude halved?

Try it yourself first, then reveal one step at a time.

Block moving right, displaced right: spring and damping both pull left.

Picture and FBD. Horizontal forces: spring −kx and damping −bv. mx″ = −kx − bx′.
Name the numbers. ω₀ = √(k/m) = 4.0 rad/s, γ = b/m = 2.0 s⁻¹.
Regime. 2ω₀ = 8.0 is bigger than γ = 2.0, so it is underdamped: it oscillates.
Frequency. ω_d = √(ω₀² − γ²/4) = √15 = 3.87 rad/s.
Half amplitude. Amplitude ∝ e^−γt/2 = e^−t, so it halves at t = ln 2 = 0.69 s.

Common mistakes

Thinking a driven oscillator settles at its own frequency. It moves at the drive frequency.
Confusing the transient with the steady state.
Assuming the peak is always exactly at ω₀.

Three kinds of damping, and resonance

The trick: complex exponentials. Guess x = e^st. Every derivative just multiplies by s, so the differential equation becomes the algebra problem s² + γs + ω₀² = 0. Complex roots mean oscillation, and Euler's formula e^iθ = cos θ + i sin θ turns them back into sines and cosines.
Underdamped (γ < 2ω₀): oscillates at ω_d = √(ω₀² − γ²/4) inside an envelope e^−γt/2. Energy decays as e^−γt.
Critically damped (γ = 2ω₀): returns fastest without overshooting. Car suspensions and door closers aim near here.
Overdamped (γ > 2ω₀): creeps back slowly.
Resonance: peak near ω₀ (slightly below it when damping is large, at √(ω₀² − γ²/2)), height about Q times the low-frequency response, width about γ. High Q means sharp and tall.
Phase: the response lags the drive by 0° at low ω, 90° at resonance, nearly 180° far above.

Check yourself: car shock absorbers are designed to be close to:

WeekCalculus

Traveling waves

A wave moves a pattern, not the stuff. Each bit of string only goes up and down, yet the shape races along.

The intuition

Think of a stadium wave: every fan just stands and sits, but the wave circles the stadium. On a string, each piece copies the motion of its neighbor a moment later.

How fast the pattern moves depends only on the string: tighter strings snap back faster, heavier strings respond slower, so v = √(T/μ). Shake faster and you don't speed the wave up; you just pack the wiggles closer together.

Assumption upgrade

Before: we tracked single objects, or rigid ones
So we didn't need: to think about a material bending
Now: a string is a continuous chain where each piece pulls on its neighbors
So we need: tension on a tiny piece of string, which leads to the wave equation

y = A sin(kx − ωt)moves toward +x

v = fλ = ω/kwave speed

v = √(T/μ)string

∂²y/∂x² = (1/v²) ∂²y/∂t²wave equation

Wave on a 4 m string

Add reflected wave Slow motion ×¼

Tension T

Mass density μ

Frequency f

Amplitude A

Wave speed v—

Wavelength λ—

k = 2π/λ—

ω = 2πf—

Max string speed—

The orange dot is one piece of string: it only moves up and down (green arrow), while the crests travel sideways at v. Turn on the reflected wave to make a standing wave, the topic of Waves II.

Try this

Predict what will happen first. Then set it up and check.

Tighten the string.
What you should see
The speed and wavelength grow; the frequency stays the same.
Shake it faster.
What you should see
The wavelength shrinks; the speed stays the same.
Add the reflected wave.
What you should see
A standing wave: some points never move (nodes).

Worked example

A 2.0 m string with mass 20 g is under 40 N of tension and shaken at 50 Hz. Find the wave speed and wavelength.

Try it yourself first, then reveal one step at a time.

A tiny piece of string: tension pulls each end along the string. Slightly different angles leave a small net vertical force.

Picture and FBD. Look at one tiny piece of string. Tension pulls on both ends, along the string. Where the string curves, the two pulls don't cancel, and that net force is what makes the piece move up and down. This leads to the wave equation and v = √(T/μ).
Mass per length. μ = 0.020/2.0 = 0.010 kg/m.
Wave speed. v = √(T/μ) = √(40/0.010) = 63 m/s.
Wavelength. λ = v/f = 63/50 = 1.26 m.
Check. Shaking faster would shorten λ but not change v. ✓

Common mistakes

Confusing wave speed with how fast a bit of string moves up and down.
Thinking a higher frequency makes the wave travel faster.
Mixing up k and λ. k = 2π/λ.

More about waves

Any shape y = f(x − vt) moves right at speed v without changing; f(x + vt) moves left. Both satisfy the wave equation.
Transverse (string, light): wiggle perpendicular to travel. Longitudinal (sound): wiggle along it.
Superposition: overlapping waves add. Waves pass through each other unchanged.
Reflection: from a fixed end the pulse comes back upside down; from a free end it comes back upright.
Energy: a sine wave on a string carries power P = ½μω²A²v.

Check yourself: you shake the same string at twice the frequency. What changes?

Week

Interference

Two waves in the same place add up. Depending on their timing, they can make each other louder or cancel out.

The intuition

Two speakers play the same note. Where crests from both arrive together, the sound is loud; where a crest meets a trough, they cancel. Which happens depends only on the difference in distance to the two speakers: a whole number of wavelengths is loud, a half-number is quiet.

Two-source interference

Show average loudness

Source spacing d

Wavelength λ

Phase shift between sources

d / λ—

Loud directions—

First loud angle off center—

Loud where d sin θ = mλ (sources in phase). Shift one source by 180° and the center line goes quiet.

Try this

Predict what will happen first. Then set it up and check.

Shift one source by half a cycle.
What you should see
The center line, which was loud, goes quiet.
Spread the sources apart.
What you should see
More loud directions, packed closer together.
Show the average loudness.
What you should see
The moving ripples disappear, leaving bright loud stripes and pale quiet ones.

Worked example

Two speakers 3.0 m apart play the same 343 Hz tone in phase. Sound travels at 343 m/s. Far away, at what angles is it first quiet, and first loud again, off the center line?

Try it yourself first, then reveal one step at a time.

No forces in this problem: interference is about how two waves add, not about pushes on an object.

Picture. Two sources, a far-away listener at angle θ. The path difference is d sin θ.
Wavelength. λ = 343/343 = 1.0 m.
First quiet spot. Half a wavelength of path difference: 3.0 sin θ = 0.5, θ = 9.6°.
First loud spot off center. 3.0 sin θ = 1.0, θ = 19.5°.
Check. Quiet comes before loud as you move off center, as it should. ✓

Common mistakes

Using path length instead of path difference.
Forgetting the half-wavelength for quiet spots.
Assuming the center is always loud. A phase shift changes that.

Path difference is everything

Loud (constructive): path difference Δr = mλ, crest meets crest.
Quiet (destructive): Δr = (m + ½)λ, crest meets trough.
Far away the path difference is d sin θ, so the loud directions are at d sin θ = mλ.
Phase shift: if one source starts half a cycle late, loud and quiet swap. Noise-cancelling headphones use exactly this.

Check yourself: two speakers play in phase. A spot is 1.5 wavelengths farther from one speaker than the other. What do you hear there?

Week

Standing waves

Trap a wave between two ends and only certain wavelengths fit. That is why a guitar string or a flute plays definite notes instead of noise.

The intuition

A guitar string is pinned at both ends, so whatever pattern it makes must have zero motion there. Only whole numbers of half-wavelengths fit: one hump, two humps, three. Each fit is a note, and together they're the harmonic series.

A pipe works the same way with sound: the air inside sloshes back and forth, and the ends decide which patterns fit. A closed end blocks the air; an open end lets it move freely.

Mode

Length L

Wave speed v

Frequency—

Wavelength—

Fundamental—

Allowed—

N marks nodes (no motion), A antinodes (most motion). For pipes the curve is the air's back-and-forth displacement. A closed end must be a node; an open end is an antinode.

f_n = nv/(2L), n = 1, 2, 3…string, open pipe

f_n = nv/(4L), n = 1, 3, 5…closed pipe

Try this

Predict what will happen first. Then set it up and check.

Step the string to n = 4.
What you should see
Four humps; the frequency is 4 times the fundamental.
Switch to a pipe closed at one end.
What you should see
Only odd harmonics, and the lowest note is half the open pipe's.
Make the pipe longer.
What you should see
Lower notes. That is why bass instruments are big.

Worked example

A guitar string 0.65 m long plays a fundamental of 110 Hz. Find the wave speed on the string and the frequency of the second harmonic.

Try it yourself first, then reveal one step at a time.

A piece of the vibrating string: unbalanced tension pulls it back toward the middle.

Picture and FBD. Both ends are fixed (nodes). The fundamental is one hump: half a wavelength fits in L. The tension on each piece pulls it back toward the straight line, which is what keeps it vibrating.
Wavelength. λ₁ = 2L = 1.30 m.
Wave speed. v = fλ = 110·1.30 = 143 m/s.
Second harmonic. Two humps, f₂ = 2f₁ = 220 Hz, one octave higher.
Check. Harmonics of a string are whole-number multiples of the fundamental. ✓

Common mistakes

Putting nodes at the open end of a pipe. Open ends are antinodes.
Using v/(2L) for a pipe closed at one end. It is v/(4L).
Thinking a longer string plays higher.

Check yourself: a pipe closed at one end can play which harmonics?

Week

Beats and the Doppler effect

Two everyday effects of waves in time: the pulsing of two almost-matching notes, and the pitch shift of a moving siren.

The intuition

Two notes slightly out of tune drift in and out of step. When they line up the sound is loud, when they are opposite it is quiet, so the loudness pulses: beats.

A siren driving toward you squashes its waves together in front, so you hear a higher pitch. As it drives away, the waves behind it get stretched out and the pitch drops.

Beats: two tones added

Tone 1

Tone 2

Beat frequency—

Average pitch—

Frequencies are scaled down so you can see individual cycles. The loudness swells |f₁ − f₂| times per second. Musicians tune by turning the peg until the beats slow to nothing.

Doppler effect: a moving source

Source speed, as a fraction of sound speed

Source speed—

Heard ahead—

Heard behind—

For a 500 Hz siren, with sound at 343 m/s. Past the speed of sound the wavefronts pile into a cone: a sonic boom. General form: f′ = f(v ± v_o)/(v ∓ v_s), with the signs chosen so approaching raises the pitch.

Try this

Predict what will happen first. Then set it up and check.

Tune the two tones to match.
What you should see
The beats vanish.
Pull them 2 Hz apart.
What you should see
Two loudness pulses per second.
Push the source past the speed of sound.
What you should see
The wavefronts pile into a cone, a shock wave.

Worked example

An ambulance with a 500 Hz siren drives toward you at 30 m/s, passes, and drives away. Sound travels at 343 m/s. What frequencies do you hear?

Try it yourself first, then reveal one step at a time.

Ambulance at constant velocity: forces balance.

Picture and FBD. Constant velocity, so the forces balance. The pitch change comes from the motion of the source through the air, not from any force.
Approaching. Waves get bunched up: f′ = 500·343/(343 − 30) = 548 Hz.
Receding. Waves get stretched: f′ = 500·343/(343 + 30) = 460 Hz.
The drop you hear. About 88 Hz as it passes.
Check. Toward you is higher, away is lower. ✓

Common mistakes

Adding the two frequencies to get the beat frequency.
Doppler sign mix-ups. Moving toward each other always raises the pitch.
Thinking the source changes the frequency it emits. It is the spacing of the waves that changes.

The formulas

Beats: f_beat = |f₁ − f₂|, at an average pitch (f₁ + f₂)/2.
Doppler: f′ = f(v ± v_o)/(v ∓ v_s). Choose the signs so moving toward each other raises the pitch.
Moving source vs moving listener: the formulas differ slightly because sound travels through air, so motion relative to the air matters.
Shock waves: when the source outruns its own sound, the wavefronts pile up into a cone, heard as a sonic boom.

Check yourself: a 440 Hz tone and a 444 Hz tone play together. How many beats per second do you hear?

Week

Fourier series

Any repeating signal can be built out of pure sine waves. That idea is behind how your ear hears pitch and how music files store sound.

The intuition

Any repeating shape is a recipe of pure sine tones: the fundamental plus harmonics at 2, 3, 4… times the frequency. Smooth shapes need only a pinch of the high harmonics. Sharp corners and jumps need lots of them. That recipe is why a violin and a flute playing the same note sound different.

Fourier series: building shapes from pure tones

Number of sine terms

Highest harmonic—

Peak overshoot—

Top: the target (dashed) and the sum so far (blue), with the first few individual sines faint. Bottom: the spectrum, how much of each harmonic goes in. Near the jump the sum always overshoots by about 9% of the jump, which is about 18% of the amplitude, no matter how many terms you add. More terms squeeze the overshoot closer to the jump but never remove it. This is called the Gibbs phenomenon.

Try this

Predict what will happen first. Then set it up and check.

Build a square wave from 40 terms.
What you should see
Sharp corners, but the overshoot at the jump never goes away.
Switch to a triangle with only 3 terms.
What you should see
Already almost perfect. Smooth shapes need few harmonics.
Try the sawtooth.
What you should see
It needs every harmonic, even and odd.

Worked example

A square wave swings between +1 and −1 at 100 Hz. List the first three nonzero harmonics and their amplitudes.

Try it yourself first, then reveal one step at a time.

No forces in this problem: Fourier series is about how a signal is built from sine waves.

Picture. A square wave is symmetric, so only odd harmonics appear: (4/π)(sin ωt + ⅓ sin 3ωt + ⅕ sin 5ωt + …).
First harmonic. 100 Hz, amplitude 4/π = 1.27.
Third harmonic. 300 Hz, amplitude 4/(3π) = 0.42.
Fifth harmonic. 500 Hz, amplitude 4/(5π) = 0.25.
Check. The fundamental is larger than the square wave itself (1.27 > 1). The higher harmonics trim its peaks down. ✓

Common mistakes

Thinking only odd harmonics ever appear. That is special to symmetric shapes like the square wave.
Expecting the overshoot to vanish with enough terms.
Confusing harmonic number with amplitude.

The recipe

A repeating signal with period T is a sum f(t) = Σ b_n sin(nωt) + a_n cos(nωt), with ω = 2π/T.
The list of amplitudes is the spectrum: how much of each harmonic goes in.
Timbre: a flute and a violin playing the same note share the fundamental but have different spectra.
Sharp corners and jumps need lots of high harmonics. Smooth shapes are mostly the fundamental.

Check yourself: which shape needs the most high harmonics to build accurately?

ReviewPHY 103

Key formulas

The main equations of the course in one place, written for this site. It is a study page, not an official exam sheet, so check what your exams allow.

Constants

g = 9.8 m/s²: near Earth
G = 6.67 × 10⁻¹¹ N·m²/kg²: gravitation

Kinematics

v = dx/dt, a = dv/dt: definitions
v_x = v_x0 + a_xt: const a
x = x₀ + v_x0t + ½a_xt²: const a
v_x² = v_x0² + 2a_x(x − x₀): const a
a_c = v²/r = ω²r: centripetal

Force and translational dynamics

ΣF = ma = dp/dt: 2nd law
|F_f| ≤ μ|F_N|: friction
F_s = −kΔx: spring
F_g = Gm₁m₂/r²: gravity
F_drag = −bv: linear drag
x_cm = Σm_ix_i / Σm_i: center of mass

Work, energy, power

W = ∫F·dr: work
K = ½mv²: kinetic
W_net = ΔK: work–energy
F_x = −dU/dx: force from U
U_s = ½kΔx²: spring
ΔU_g = mgΔy: near surface
U_G = −Gm₁m₂/r: general
P = dW/dt = F·v: power

Linear momentum

p = mv: momentum
J = ∫F dt = Δp: impulse

Rotation

ω = dθ/dt, α = dω/dt: definitions
ω = ω₀ + αt: const α
θ = θ₀ + ω₀t + ½αt²: const α
ω² = ω₀² + 2α(θ − θ₀): const α
v = rω, a_t = rα: tangential
τ = r × F, |τ| = rF sin θ: torque
Στ = Iα: 2nd law
I = ∫r² dm = Σmr²: moment of inertia
I = I_cm + Md²: parallel axis
K = ½Iω²: rotational KE
L = Iω, L = r × p: angular momentum
ΔL = ∫τ dt: angular impulse
W = ∫τ dθ: rotational work

Oscillations

x = A cos(ωt + φ): SHM
T = 2π/ω = 1/f: period
T_s = 2π√(m/k): spring
T_p = 2π√(ℓ/g): pendulum
T = 2π√(I/mgd): physical pendulum

Geometry and calculus

C = 2πr, A = πr²: circle
V = ⁴⁄₃πr³, S = 4πr²: sphere
d(xⁿ)/dx = nxⁿ⁻¹: power rule
d(e^ax)/dx = ae^ax: exponential
d(ln ax)/dx = 1/x: log
d(sin ax)/dx = a cos ax: sine
d(cos ax)/dx = −a sin ax: cosine
∫xⁿ dx = xⁿ⁺¹/(n+1): n ≠ −1
∫dx/(a + bx) = (1/b) ln|a + bx|: log form
A·B = AB cos θ: dot
|A × B| = AB sin θ: cross

Waves

v = fλ = ω/k: wave speed
v = √(T/μ): string
f_n = nv/(2L): string, open pipe
f′ = f(v ± v_o)/(v ∓ v_s): Doppler
f_beat = |f₁ − f₂|: beats

ReviewPHY 103

Which principle do I use?

Most of the time lost on exams goes into choosing a method. Learn the clues, then practice picking.

The intuition

Each big principle answers a different kind of question. Forces tell you how things accelerate right now. Energy connects "before" and "after" without caring about the path. Momentum handles quick, violent interactions where the forces are unknown. Angular momentum does the same for spinning. Torque balance handles things that stay put.

Many hard problems are two easy problems glued together. Split them at the moment the physics changes (a collision ends, a string goes slack).

Clue in the problem	Reach for
Asks for a time, or gives a constant acceleration	Kinematics (plus forces to find a)
Asks for a force: tension, normal force, friction	Forces: free-body diagram and ΣF = ma (ΣF = mv²/r on curves)
Asks for a speed after a height or spring change; the path is curvy or unknown	Energy
A collision, explosion, or push-off; a very brief, huge force	Momentum (impulse for the force)
Something spinning changes shape, or something sticks to a pivoted or spinning object	Angular momentum
An extended object at rest with hinges, cables, supports	Torque balance Στ = 0 (and ΣF = 0)
A collision followed by a swing or slide	Momentum for the collision, then energy
A force at a specific point on a curved path (top of a loop, bottom of a swing)	Energy for the speed, then forces for the force

Practice: pick the principle

Choose the first tool you would reach for.

1. A car brakes from 30 m/s to rest in 4.0 s. How far does it travel?

2. A skier starts from rest at the top of a 40 m frictionless hill of unknown shape. How fast are they going at the bottom?

3. Two skaters at rest push off each other. One moves away at 3 m/s. How fast does the other move?

4. A sled is pulled at constant speed up a rough slope. What is the rope tension?

5. A spinning skater pulls in their arms. What is the new spin rate?

6. A uniform beam is held by a hinge at one end and a cable at the other. What is the cable tension?

7. A bullet embeds in a hanging wooden block. How high does the block swing?

8. A block slides down a rough ramp. How long does it take to reach the bottom?

9. A spring launches a ball straight up. How high does it go?

10. A ball rolls without slipping down a ramp. How fast is it at the bottom?

11. A lump of clay hits and sticks to the end of a rod pivoted at its other end. How fast does the rod spin right after?

12. A hockey puck hits the boards and bounces back. What average force did the boards exert during the 0.01 s contact?

13. A pendulum bob is released from 30°. What is the string tension at the bottom?

14. A satellite's elliptical orbit brings it as close as 7000 km to Earth's center, where it moves at 8.7 km/s. How fast is it at the far point, 14,000 km from the center?

15. A ball is thrown horizontally off a cliff. Where does it land?

16. You push a merry-go-round with a known torque. How long until it reaches 2 rad/s?

ReviewPHY 103

Exam review

Build a mixed problem set from the weeks your midterm or final covers. No hints during the set; everything is graded and explained when you submit.

Which weeks does your midterm or final cover?

Number of problems

Timer

ReviewSpaced practice

Review my mistakes

Every problem you miss in the practice set or exam review lands here and comes back with new numbers. Spacing reviews out over days is one of the most reliable ways to make things stick.

How the queue works

Miss a problem: it's due right away. Get it right: it comes back in 1 day, then 3 days, then 7. After three correct reviews in a row it's mastered and leaves the queue. Miss it again and the count restarts.

All weeks

Practice set

problems with randomized numbers, covering every week of the course. Answers within 2% count. Use g = 9.8 m/s². Progress is saved in this browser.

0 of 15 solved

Mechanics, with the calculus left in.

New to physics? Start here

How to use a module

Five ideas you'll use everywhere

Suggested path

Modules by week

How this maps to your course

Vectors

Try this

Worked example

Common mistakes

The toolkit

Derivatives: how fast things change

Try this

Worked example

Common mistakes

Rules you will use

Integrals: adding up

Try this

Worked example

Common mistakes

Rules you will use

Dot and cross products, step by step

Try this

Worked example

Common mistakes

Unit vectors

Differential equations you will meet

Try this

Worked example

Common mistakes

The equations in this course

Motion graphs

Try this

Worked example

Common mistakes

Reading the graphs

Constant acceleration

Try this

Worked example

Common mistakes

Where the equations come from

Drag and changing acceleration

Assumption upgrade

Try this

Worked example

Common mistakes

Worked case: falling with linear drag

Two more patterns to recognize

Projectiles

Simplification for now

Try this

Worked example

Common mistakes

Things exams test

Relative motion

Reference frames: who is measuring?

Try this

Worked example

Common mistakes

The two classic questions

Circular motion

Assumption upgrade

Try this

Worked example

Common mistakes

Deriving ac with calculus

Curves and fictitious forces

Assumption upgrade

Try this

Worked example

Common mistakes

How to solve curve problems

Forces and friction

Assumption upgrade

Try this

Worked example

Common mistakes

A recipe for any force problem

Which way does friction point?

Deriving a_c with calculus