Lessons

A ball leaves your hand at 20 m/s straight up, from a height of 1.5 m. How high does it go, when does it land, and how fast is it moving when it hits the ground?

A ball is launched from the ground at $v_0 = 20$ m/s, $\theta = 40°$ above the horizontal (no air resistance, $g = -10$). What is its range, how high does it go — and why is the path a parabola? Drag the launch angle and speed and watch the trajectory, range, and apex move together.

A block is released from rest on a $\theta = 30°$ incline with kinetic friction $\mu = 0.2$. Resolving forces along the slope gives a constant acceleration $a = g(\sin\theta - \mu\cos\theta)$ — and crucially, the mass cancels. From there the motion is constant-acceleration kinematics: $v = at$ and $x = \tfrac12 at^2$, which are just the integrals of that constant $a$. Drag the angle and the friction coefficient and watch all three panels steepen together.

An object is dropped from rest and feels air resistance proportional to its speed (m = 1 kg, drag b = 0.5 kg/s). Does it keep speeding up forever — and why can't the algebra formulas answer that?

The same launch — $v_0 = 30$ m/s at $\theta = 45°$ — but now with quadratic air resistance $F = b\,|v|\,v$. There is no closed-form trajectory: the path is solved by numerical integration. Slide the drag up from zero and watch the clean parabola (dashed) deform — the range shrinks and the descent steepens.

A 2 kg block on a frictionless track starts at rest. A force that grows with distance, $F(x) = b\,x$ (here $b = 8$ N/m), pushes it for $d = 1.5$ m. How much work is done — and where does $\tfrac12 m v^2$ actually come from? Watch the work accumulate as the area under the force curve.

A $m = 2$ kg block is released from rest at the top of a frictionless ramp, $H = 20$ m above the bottom. As it slides down, potential energy turns into kinetic energy — but the total never changes. How fast is it going at the bottom, and would a steeper ramp change that? Drag the height: watch the KE and PE bars trade while the Total bar stays flat. The speed at any height depends only on how far it has dropped, not on the shape of the ramp.

A $0.5$ kg ball is struck by a bat: the contact force rises and falls as a brief pulse, $F(t) = F_{\max}\sin(\pi t/\tau)$ with $F_{\max} = 200$ N over $\tau = 0.02$ s. What impulse does it deliver, and how fast does the ball leave? Watch the impulse accumulate as the area under the force–time curve.

A $m_1 = 2$ kg cart moving at $v_1 = 3$ m/s strikes a stationary $m_2 = 1$ kg cart. Slide the coefficient of restitution $e$ from $1$ (perfectly elastic — the carts bounce apart with no energy lost) down to $0$ (perfectly inelastic — they latch together). The momentum bars are the same total height before and after at every setting; the kinetic-energy bars match only when $e = 1$, and the gap that opens as $e \to 0$ is energy lost to deformation. Momentum is conserved because the contact forces are equal and opposite (Newton's third law); kinetic energy is not, because the deformation work need not come back.

A flywheel starts from rest and spins up under a constant angular acceleration $\alpha = 3$ rad/s² for $4$ s. How fast is it turning, and how many revolutions has it made? The rotational kinematics are the straight-line equations with angular symbols — step them and watch the timeless equation fall out of the integrals.

A uniform rod of mass $M = 2$ kg and length $L = 1.2$ m spins about one end. The algebra course just hands you $I = \tfrac13 ML^2$ from a table — but where does the $\tfrac13$ come from? Each slice at radius $r$ adds $r^2\,dm$, so the moment is the area under $dI/dr = \lambda r^2$. Drag the cursor out from the pivot and watch $I$ build up — slowly near the axis, fast near the tip.

A flywheel (moment of inertia $I = 0.5$ kg·m²) starts from rest and is driven by a torque that builds with angle, $\tau = c\theta$ ($c = 8$ N·m/rad), through $\theta = 1.5$ rad. How fast is it spinning at the end? The work done is the area under the torque–angle curve — and it is exactly the rotational kinetic energy $\tfrac12 I\omega^2$. Drag the cursor and watch the shaded work and the spin energy grow together: the rotational twin of $\int F\,dx = \tfrac12 mv^2$.

A 1 kg mass on a spring (k = 4 N/m) is pulled 0.3 m from equilibrium and released from rest. How does it move — and where does the period the algebra course made you memorize actually come from?

Add friction to the mass on a spring. As the damping grows the motion changes character — oscillating, then just barely returning, then crawling back. Where is the boundary, and what is the fastest a system can return to rest?

A satellite circles a planet ($\mu = GM = 4\times10^{14}$ m³/s²) at radius $R = 7\times10^{6}$ m. Gravity pulls it straight toward the centre the whole time — so why doesn't it fall in? Because at orbital speed the inward pull is exactly the centripetal acceleration a circle needs: the satellite is in continuous free fall, falling around the planet. Drag the radius outward and watch the orbit widen while the speed drops as $1/\sqrt{R}$ and the period stretches as $R^{3/2}$ — Kepler's third law, live.

A real orbit is not a circle but an ellipse with the planet at one focus ($\mu = GM = 4\times10^{14}$ m³/s², semi-major axis $a = 1.6\times10^{7}$ m). The satellite races through its closest point (perihelion) and crawls at its farthest (aphelion). There is no algebra formula for where it is at time $t$ — so the path is integrated numerically from $\ddot{\mathbf r}=-\mu\mathbf r/r^3$, and SymPy checks it conserves energy and angular momentum. Slide the eccentricity from a circle to a long ellipse: the shape changes completely, but because every orbit shares the same semi-major axis, they all take the same time to go around — Kepler's third law.

Lifting a mass away from a planet ($\mu = GM = 4\times10^{14}$ m³/s², surface radius $R = 6.4\times10^{6}$ m): the gravitational pull weakens as $F = GMm/r^2$. How much energy does it take to climb — and why is the energy to escape finite? Drag the cursor outward and watch the potential energy approach a ceiling as the area under $1/r^2$ converges.

A vertical wall $w = 5$ m wide holds back fresh water ($\rho = 1000$ kg/m³) to a depth of $H = 4$ m. The water pressure grows with depth as $P = \rho g h$, so the wall is pushed hardest at the bottom and not at all at the surface. What is the total force on the wall? It is the area under the pressure curve — drag the cursor to raise the water level and watch the force grow as the square of the depth.

Water drains from an orifice a depth $h = 5$ m below the surface of a tank. A parcel of water falling from the surface trades potential energy for kinetic energy, and leaves the orifice at $v = \sqrt{2gh}$ — exactly the speed it would reach falling freely through $h$. Drag the cursor down from the surface: the KE and PE bars (per unit volume) trade while the Total bar stays flat. The efflux speed is set by the depth alone, not by the width of the tank or how much water sits above.

One mole of an ideal gas at $T = 300$ K expands isothermally from $V_1 = 0.025$ m³ to $V_2 = 0.075$ m³. How much work does it do — and why is the answer a logarithm, not simply pressure times volume change? Watch the work accumulate as the area under the P–V curve.

A diatomic gas ($\gamma = 1.4$) expands from $V_1 = 1$ L at $P_1 = 100$ kPa to $V_2 = 3$ L with no heat exchanged ($PV^\gamma =$ const). How much work does it do, and what happens to its temperature? The work is again the area under the $P$–$V$ curve — but the adiabat falls off faster than an isotherm, so the gas does less work and cools. Drag the cursor out along the adiabat to watch the work accumulate; raise the initial pressure $P_1$ to lift the whole curve while its shape (and the cooling ratio) stays fixed by $\gamma$.

Two opposite point charges, of magnitudes $q_1 = q_2 = 2\ \mu$C, are held a separation $R = 2$ cm apart and attract with $F = kq_1q_2/r^2$. How much energy does it take to pull them fully apart — and why is that energy finite? Drag the cursor outward and watch the potential energy climb toward a ceiling as the area under $1/r^2$ converges. This is the electric twin of escape energy: the same inverse-square area, charge in place of mass.

A rod of length $L = 0.3$ m with linear charge density $\lambda = 100$ nC/m lies along the axis, its near end $a = 0.1$ m from the point where we want the field. Algebra gives the field of a point charge, $kq/r^2$ — but a rod is not a point, and there is no algebra formula for charge spread along a length. Calculus is the only way in: slice the rod into point charges $dq = \lambda\,dx$, each contributing $k\,dq/x^2$, and integrate. The shaded area under those contributions is the field $E = k\lambda(1/a - 1/(a+L))$, which collapses to the point-charge $kQ/a^2$ when the rod is short.

A $C = 500\ \mu$F capacitor is charged until it holds $Q = 0.1$ C, reaching $V = 200$ V. How much energy is stored — and why is it $\tfrac12 CV^2$ and not $CV^2$? As charge piles on, the voltage climbs as $V = q/C$; the energy is the area under that rising line. Drag the cursor and watch the stored energy grow as the triangle under $V(q)$.

A capacitor $C = 100\ \mu$F is charged to $Q_0 = 0.01$ C and then connected across an inductor $L = 10$ mH. With no resistance to drain it, the charge $Q(t)$ and the current $I(t)$ oscillate forever at angular frequency $\omega = 1/\sqrt{LC}$. Plotted as a stacked $Q$–$t$ over $I$–$t$, the slope of the charge is the current at every instant — the same slope↔value pivot as position and velocity, and the electrical twin of a mass on a spring. Drag $L$ and watch the period $T = 2\pi\sqrt{LC}$ stretch; the energy sloshes between the capacitor and the inductor while the total stays fixed.

A coil of area $A = 0.1$ m² spins at $\omega = 10$ rad/s in a uniform magnetic field $B = 0.5$ T. The flux through it varies as $\Phi(t) = BA\cos(\omega t)$, and the induced EMF is its rate of change, $\mathcal{E} = -d\Phi/dt = BA\omega\sin(\omega t)$ — a sinusoid that lags the flux by a quarter cycle (its peak comes a quarter turn after the flux's). This is how a generator turns rotation into alternating current. Plotted as a stacked $\Phi$–$t$ over EMF–$t$, the slope of the flux is the (negated) EMF at every instant. Drag $B$ or $\omega$: when the flux is at its peak the EMF is zero, and when the flux crosses zero the EMF is at its peak.

A diverging lens (concave — thinner in the middle) with focal length $f = -1$ m. The focal length is negative because parallel rays leave it spreading apart, as if they had come from a focus in front of the lens. An object $h_o = 0.4$ m tall stands a distance $d_o$ in front. Here is the surprise: no matter where you put the object — right up close or far away — a diverging lens makes only one kind of image: virtual, upright, and shrunken, on the same side as the object. Slide the object and watch — the image never flips, never enlarges, never lands on a screen. This is the lens in a door peephole and the front group of a wide-angle view.

A disk of radius $R = 0.3$ m carries a uniform surface charge density $\sigma = 100$ nC/m². We want the electric field a distance $z = 0.1$ m out along its axis. Algebra hands you two extremes — a point charge $kQ/z^2$ and an infinite sheet $\sigma/2\varepsilon_0$ — but a real disk is neither, and there is no algebra formula for one. Calculus is the only way in: slice the disk into concentric rings $dq = \sigma\,2\pi r\,dr$, each contributing $k\,dq\,z/(z^2+r^2)^{3/2}$ on the axis, and integrate. The shaded area under those ring contributions is the field $E = \tfrac{\sigma}{2\varepsilon_0}(1 - z/\sqrt{z^2+R^2})$ — which collapses to the point charge far away and to the infinite sheet for a huge disk.

A $m_1 = 1$ kg block moving right at $v_1 = +4$ m/s meets a $m_2 = 2$ kg block moving left at $v_2 = -2$ m/s, head-on. Their momenta are equal and opposite, so the total momentum is exactly zero. Slide the coefficient of restitution $e$ from $1$ (perfectly elastic — the blocks bounce apart) down to $0$ (perfectly inelastic — they latch together): the momentum total stays pinned at zero the whole way (only the split between the two bodies flips, above and below the line), while the kinetic-energy total collapses — at $e = 0$ the pair comes to a dead stop and all $12$ J is lost. Zero total momentum is still a conserved momentum; it does not mean the collision does nothing.