Y12 → Y13 Bridge / Paper 3 Data Hunt

Paper 3 Data Hunt

Paper 3 is "easily fraudable" once you spot the pattern: either every value you need is already given, or you're expected to know the missing equation from the formula booklet. For each scenario, sort the cards into the right category.

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Satellite in Circular Orbit

A satellite moves in a circular orbit of radius $7.0\times10^{6}\ \text{m}$ around the Earth (mass $M=5.97\times10^{24}\ \text{kg}$, $G=6.67\times10^{-11}\ \text{N m}^2\text{kg}^{-2}$). Find the satellite’s orbital speed.

Tap a card below, then tap the category it belongs to.

Orbital radius r = 7.0×10⁶ m
Earth's mass M = 5.97×10²⁴ kg
G = 6.67×10⁻¹¹ N m² kg⁻²
F = GMm/r² — Newton's law of gravitation
F = mv²/r — centripetal force
T = 2π√(l/g) — simple pendulum period

Discharging Capacitor

A $470\ \mu\text{F}$ capacitor charged to $6.0\ \text{V}$ discharges through a $22\ \text{k}\Omega$ resistor. Find the voltage remaining after $5.0\ \text{s}$.

Tap a card below, then tap the category it belongs to.

Capacitance C = 470 μF
Resistance R = 22 kΩ
Initial voltage V₀ = 6.0 V
Time t = 5.0 s
V = V₀e^(-t/RC) — exponential decay of voltage
Q = CV — charge stored, not asked for here

Oscillating Mass on a Spring

A mass on a spring oscillates with amplitude $0.050\ \text{m}$ and angular frequency $4.0\ \text{rad s}^{-1}$. Find the speed of the mass when its displacement is $0.030\ \text{m}$.

Tap a card below, then tap the category it belongs to.

Amplitude A = 0.050 m
Angular frequency ω = 4.0 rad s⁻¹
Displacement x = 0.030 m
v = ±ω√(A²-x²) — SHM speed-displacement equation
x = Acos(ωt) — needs a time value, which isn’t given
T = 2π√(m/k) — period of a mass-spring system, not asked for

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