§ 00 · Visual overview

See mechanics.
Don't just memorize it.

Every core idea of classical mechanics — motion, force, energy, and waves — lives here as a simulation you can grab, drag, and replay. The equations aren't the physics; they're the caption. The picture is the physics.

ball path velocity vector A ball bouncing with energy loss — kinematics, forces, and energy in one frame. It restarts on its own.
§ 01 · Kinematics

Projectile motion is two problems in one

Gravity only pulls down — so the horizontal motion never feels it. Split the launch velocity into components and each axis becomes a problem you already know how to solve: constant velocity across, constant acceleration up-and-down.

x(t) = v₀ cos θ · t     y(t) = v₀ sin θ · t − ½gt²
trajectory velocity v vₓ (constant!) vᵧ (changing) Drag the launch arrow, or use the sliders.
Range
–m
Max height
–m
Flight time
–s
v at impact
–m/s

Watch the green arrow as it flies: it never changes. Only the red vertical component grows and shrinks. Try θ = 30° and θ = 60° — same range. Complementary angles always land in the same spot.

§ 02 · Forces & Newton's laws

A force diagram is an argument

Will the block slide? Don't guess — decompose. Tilt gravity's pull into a piece along the ramp and a piece into it. Friction can only fight back up to its limit μN. The moment the along-ramp pull wins, the block moves.

slides when  mg sin θ > μ · mg cos θ  ⇔  tan θ > μ
weight mg mg sin θ (pull) normal N friction f
Pull along ramp
–N
Max static friction
–N
Verdict
–
Acceleration
–m/s²

Notice mass never decides the verdict — it cancels from both sides. The tipping point is pure geometry: the block slips exactly when tan θ = μ. Slide θ up slowly and watch the friction arrow max out, then lose.

§ 03 · Energy conservation

Energy doesn't vanish — it changes costume

Lift a pendulum and you've stored energy as height. Release it and that energy trades into speed, then back, forever (minus a little tax to air drag). The two bars below always sum to the same total — that's the whole law.

½mv² + mgh = Etotal = constant
kinetic ½mv² potential mgh Drag the bob to any angle, then let go.
Kinetic
–J
Potential
–J
Total
–J
Speed at bottom
–m/s

With drag off, the total bar is frozen — the definition of conservation. Turn drag on and watch the total bleed away while the swing shrinks to match. Energy accounting never lies.

§ 04 · Waves & superposition

Waves don't collide — they add

Two waves passing through each other simply sum, point by point, instant by instant. In phase they reinforce; half a cycle apart they can erase each other completely. That single fact powers noise-cancelling headphones, interference fringes, and beats.

y(x,t) = y₁ + y₂ = A₁ sin(k₁x − ω₁t) + A₂ sin(k₂x − ω₂t + φ)
wave 1 wave 2 sum

Hit Destructive: identical waves, φ = 180°, and the black sum flatlines — two waves making silence. Then try Beats: nearly-equal frequencies drift in and out of phase, so the sum throbs. That throb is what you hear tuning a guitar.

§ 05 · Practice

Now you do the physics

Each problem is solvable with one idea from above. Hints reveal one step at a time — try before you peek, and use the simulators to check your answer.

1 · The cliff throw

A ball is thrown horizontally at 15 m/s from a 20 m cliff. How far from the base does it land? (g = 9.8 m/s²)

Hint 1 — The two axes are independent. Vertically it's just a drop: 20 = ½ · 9.8 · t².
Hint 2 — Solve for time: t = √(2·20/9.8) ≈ 2.02 s. The horizontal axis just coasts for that long.
Answer — x = 15 × 2.02 ≈ 30.3 m. Notice the throw speed never entered the falling time.

2 · The stubborn crate

A crate sits on a ramp with μ = 0.4. At what angle does it start to slide?

Hint 1 — Sliding starts when the pull along the ramp equals maximum friction: mg sin θ = μ · mg cos θ.
Hint 2 — Mass cancels. Divide both sides by mg cos θ and you're left with tan θ = μ.
Answer — θ = arctan(0.4) ≈ 21.8°. Verify it on the § 02 simulator: set μ = 0.40 and creep the angle past 21°.

3 · The swing speed

A pendulum bob is released from a height 0.8 m above its lowest point. How fast is it moving at the bottom?

Hint 1 — No equations of motion needed — this is pure energy bookkeeping: mgh = ½mv².
Hint 2 — Mass cancels again (spot the theme?): v = √(2gh).
Answer — v = √(2 · 9.8 · 0.8) ≈ 3.96 m/s. Drag the § 03 bob up and compare with the “speed at bottom” readout.