Pick your class, read the steps, then try them right in the sandbox —
change one slider at a time and watch the graphs respond.
Class 9 — Equations of motion & free fall
Concept: under gravity alone, a dropped object accelerates uniformly:
v = gt and s = ½gt² (starting from rest). The velocity-time graph should
be a straight line through the origin — constant acceleration — and the
distance-time graph should curve, since distance grows with the square of
time.
- Switch to Free fall mode.
- Set Drop height to 20 m and leave Gravity at 9.8 m/s².
- Press ▶ Play and watch the v-t graph — check it really is a
straight line.
- Reset, then pick Jupiter (about 24.8 m/s²) from the planet menu and
play again. The object reaches the ground faster — read the new time off the graph
and compare it with t = √(2h/g). Try the Moon too.
- Tick compare two masses, set Mass A to about 1 kg and Mass B to 80 kg, and
play. They land together, exactly as the formula predicts.
- Now tick air resistance and play again. The heavier body wins, because drag
slows a light body more. That's why a feather only falls like a hammer in a
vacuum.
Takeaway: the time to fall a fixed height depends only on height and
gravity, never on mass — that's why a heavy and light object dropped together land
together (no air resistance).
Class 11 — Projectile motion: range and time of flight
Concept: range R = u²sin(2θ)/g and time of flight
T = 2u sin(θ)/g. Range is largest at θ = 45°, and two angles that add
to 90° (like 30° and 60°) give the same range.
- Switch to Projectile motion mode.
- Set Launch speed to 20 m/s and Angle to 45°. Press
▶ Play and note the range shown above the stage.
- Reset, change the angle to 30°, keep the same speed, and play again. Note
the (smaller) range.
- Reset, try 60° instead. Compare this range with the 30° one — they
should match, confirming the "complementary angles" rule.
Takeaway: maximum range happens at 45°, and range is symmetric around
it.
Class 11 — Simple harmonic motion: the pendulum period
Concept: for small swings, a pendulum's period doesn't depend on how far
you pull it back (amplitude) — only on its length: T = 2π√(L/g).
- Switch to Pendulum & SHM mode.
- Set Length L to 1 m and Start angle to 10°. Play and time (or
read) one full swing.
- Reset, change the Start angle to 25° only — keep the length the same — and
play again. The period should barely change, even though the swing is bigger.
- Now reset the angle to 10° and instead double the Length to 2 m. Play again
— this time the period noticeably increases, matching T ∝ √L.
Takeaway: length changes the period; starting angle (amplitude) barely
does, as long as the swing stays small.
Class 11 — SHM: restoring force and energy (JEE)
Concept: motion is simple harmonic when the restoring force is proportional to
the displacement and opposite to it: F = −kx, so a = −ω²x. For a
pendulum the restoring force is the part of the weight along the arc,
F = −mg sinθ ≈ −mgθ for small θ, which gives
ω² = g/L.
- In Pendulum & SHM tick show forces on the bob, set the start angle to
10° and Play. The orange mg sinθ arrow always points back to the centre and
flips sides as the bob crosses the bottom.
- Set the right-hand graph to Restoring force vs angle. At 10° the solid line
and the dashed mgθ line coincide. Raise the start angle to 85°: the curve bends
below the line, so the motion is no longer exactly SHM.
- Switch back to Energy vs time. KE + PE stays flat (damping 0). Now set damping
to 0.3 and watch the total drain away.
- Change the mass from 1 kg to 5 kg. The period does not change: the force grows with
m, but so does the inertia.
- Set the start angle to 60°, L = 2 m, Earth. At the lowest point the tension is
T = mg + mv²/L = 2mg = 19.6 N, not just mg.
Worked check: a 1 m pendulum on the Moon (g = 1.62 m/s²) has
T = 2π√(1/1.62) ≈ 4.94 s.
Takeaway: the string tension is largest at the bottom, the restoring force is
largest at the ends, and T is independent of mass (and of amplitude, only for small
angles).
Class 11 — Spring–mass system and spring combinations (JEE)
Concept: a mass on a spring obeys F = −kx, so T = 2π√(m/k) and
ω = √(k/m). Springs in parallel add (k = k₁ + k₂); springs in
series combine like resistors in parallel (1/k = 1/k₁ + 1/k₂).
- Open Spring–mass SHM, horizontal, m = 1 kg, k = 20 N/m. The period is
2π√(1/20) ≈ 1.40 s. Change the amplitude: the waves get taller but not
faster.
- Double the mass: T grows by √2 (≈ 1.99 s). Quadruple k instead: T is
halved.
- Choose Two springs in parallel with k₁ = 20 and k₂ = 30 N/m. The
lab shows k = 50 N/m and T ≈ 0.89 s.
- Choose Two springs in series with the same values: k = 20·30/50 =
12 N/m and T ≈ 1.81 s. Now set both to 40 N/m: k = 20 N/m, half of one spring.
- Choose Vertical. The block now oscillates about a lower equilibrium, shifted by
e = mg/k = 0.49 m. Switch the planet to the Moon: e shrinks but the period does not
change.
- Raise the damping slider and watch the amplitude decay faster and faster.
Takeaway: gravity changes where a vertical spring oscillates, never how fast.
A spring cut in half has twice the stiffness (k ∝ 1/length), because each half is
one of two springs in series.
Class 11 — SHM as the shadow of circular motion (JEE)
Concept: if P moves uniformly round a circle of radius A at angular speed
ω, its projection Q on a diameter performs SHM:
x = A cos(ωt + φ), v = −Aω sin(ωt + φ),
a = −ω²x.
- Open SHM & circular motion, A = 1.5 m, ω = 2 rad/s. The period is
2π/ω = 3.14 s, the maximum speed Aω = 3.00 m/s and the maximum
acceleration Aω² = 6.00 m/s².
- Play with the vectors on. Q is fastest at the centre (where a = 0) and momentarily at
rest at the ends (where |a| is largest).
- Set the phase φ to 90°: Q starts at the centre moving left, and the x–t
graph is −A sin ωt.
- Tick also project onto the y-axis. The two shadows are the same wave a
quarter-cycle apart.
- In the panel, check that v² + ω²x² stays equal to (Aω)²
at every instant.
Worked trick: time for Q to go from x = A/2 to x = A. Here cosθ = 1/2, so
the phase covers π/3 and t = π/(3ω) (0.52 s for ω = 2 rad/s). Phasor
angles turn most SHM timing questions into one-line arithmetic.
Class 11 — Collisions and conservation of momentum
Concept: total momentum (m1v1 + m2v2)
is conserved in every collision, but kinetic energy is only conserved when the
collision is elastic (Restitution e = 1).
- Switch to Collisions mode.
- Set Mass 1 = 2 kg, Mass 2 = 2 kg, v₁ = 5 m/s, v₂ = 0 m/s, and
Restitution e = 1.00 (elastic). Play — with equal masses, the moving object should
stop and the stationary one takes off at 5 m/s.
- Reset, set Restitution e to 0.00 (perfectly inelastic) and play again — the two
objects now move off stuck together at a shared, slower velocity.
Takeaway: momentum is conserved either way, but the two masses end up with
different final velocities depending on how elastic the collision is.