The A2 core practicals flashcards

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Impulse
Force × time for which it acts (FΔt); equal to the change of momentum. It is the area under a force–time graph.
Time constant of an RC circuit
RC: the time for the p.d. (or charge) of a discharging capacitor to fall to 1/e (about 37%) of its starting value.
Specific latent heat
The energy needed to change the state of 1 kg of a substance without a change of temperature (ΔE = LΔm).
Half-value thickness
The thickness of absorber that halves the corrected count rate: x½ = ln 2/μ.
Resonance
Large-amplitude oscillation when the driving frequency equals the natural frequency of the system.
Control (in CP13)
An identical funnel of ice with no heater, run for the same time, to measure the ice melted by the surroundings.
CP9: force and momentum
F = Δp/Δt; impulse FΔt = Δp = m(v − u). (Yes)
CP11: discharge in straight-line form
ln V = ln V₀ − t/RC (from V = V₀e^(−t/RC)). (Yes)
CP13: latent heat with a control
L = VIt / (m_heater − m_control). (No — derive it)
CP14: ideal gas
pV = NkT; at constant T, p ∝ 1/V. (Yes)
CP15: absorption by lead
C = C₀e^(−μx), so ln C = ln C₀ − μx. (No — given in the question when needed)
CP16: straight-line form for a mass–spring system
1/f² = T² = 4π²m/k (from T = 2π√(m/k)). (No — derive it)
CP9 force sensor: what gives Δp?
The area under the force–time graph from the data logger.
CP11: what to plot and what the gradient gives
ln V against t: straight line, gradient −1/RC, intercept ln V₀.
CP12: shape of the thermistor calibration graph
A smooth curve of V_out against θ, not a straight line; steepest (most sensitive) near the middle of the range.
CP14: what to plot and what the gradient gives
p against 1/V: straight line through the origin, gradient NkT.
CP15: what to plot and what the gradient gives
ln C (background-corrected) against thickness x: straight line, gradient −μ.
CP16: meaning of the small intercept on 1/f² against m
The spring's own effective mass, which adds to every load. Read the unknown mass from the line, intercept included.
CP9: why tilt the runway?
So a component of the weight cancels friction; the hanging weight is then the resultant force.
CP10: why must the scale be in the plane of the spheres?
The camera magnifies objects at different distances differently; a scale nearer the camera would give wrong distances.
CP10: how do you test momentum conservation in a glancing collision?
Find velocity components along and perpendicular to the original direction; check total momentum in each direction separately.
CP11: when is an oscilloscope better than a data logger?
When RC is small (milliseconds): the logger samples too slowly. A square wave makes the trace repeat so it can be read.
CP12: how is the fixed resistor chosen for a thermostat?
Equal to the thermistor's resistance at the set temperature, so the output is half the supply there and the divider is most sensitive.
CP14: why wait after compressing the gas?
Compression does work on the gas and warms it; waiting lets it return to room temperature.
CP15: why put thin aluminium over the source?
It absorbs any beta particles, so only gamma reaches the GM tube.
A triangular F–t pulse peaks at 30 N and lasts 0.040 s. Impulse?
Area = ½ × 30 × 0.040 = 0.60 N s = change of momentum. (No — learn the area rule (FΔt = Δp is listed))
A 0.500 kg trolley hits a buffer at 0.80 m s⁻¹ and rebounds at 0.60 m s⁻¹. Change of momentum?
Velocity reverses: Δp = 0.500 × (0.60 + 0.80) = 0.70 kg m s⁻¹. (Yes)
ln V against t has gradient −0.025 s⁻¹; R = 33 kΩ. Capacitance?
RC = 1/0.025 = 40 s; C = 40 ÷ 33 000 = 1.2 × 10⁻³ F (1200 μF). (Yes)
6.0 V across a thermistor and a 10.0 kΩ resistor in series; V across the resistor is 4.0 V. Thermistor resistance?
Thermistor has 2.0 V; same current, so R = 10.0 kΩ × 2.0/4.0 = 5.0 kΩ. (No — derive it)
Heater 12.0 V, 2.0 A for 300 s; 30.0 g melts with heater, 8.5 g in the control. L?
L = 7200 J ÷ 0.0215 kg = 3.3 × 10⁵ J kg⁻¹. (Yes)
p against 1/V has gradient 2.50 J for air at 300 K. Number of molecules?
N = gradient/kT = 2.50 ÷ (1.38 × 10⁻²³ × 300) = 6.04 × 10²⁰. (Yes)
ln C against x has gradient −0.050 mm⁻¹. Half-value thickness?
μ = 0.050 mm⁻¹; x½ = ln 2 ÷ 0.050 = 13.9 mm. (No — derive it)
1/f² against m has gradient 0.400 s² kg⁻¹. Spring constant?
k = 4π² ÷ 0.400 = 98.7 N m⁻¹. (No — derive it)
Uncertainty in a mass found by subtracting two readings
Add the absolute uncertainties (±0.1 g + ±0.1 g = ±0.2 g), then convert to a percentage of the difference.
Uncertainty in a product or quotient (A2)
Add the percentage uncertainties; for a power, multiply the percentage by the power.
Does the result agree with the accepted value?
Compare the percentage difference with the percentage uncertainty: agreement if the difference is smaller.
Forgetting the background in CP15
Subtract the background rate before taking logs, or the ln graph curves upwards at large thicknesses.
Safety with a radioactive source
Tongs, arm's length, never pointed at anyone, short exposure, back in the lead box; wash hands after handling lead.
Labelling a log axis
Take the log of a pure number: ln (V / V) or ln (C / s⁻¹). The log has no unit.

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