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Edexcel A-Level Physics: Synoptic & Practical Skills

6 exam-style questions with full mark schemes and model answers. Write your own answer and the AI examiner marks it against the mark scheme.

Question 19 marksExplain

A satellite moves in a circular orbit around the Earth. The behaviour of such a satellite links two areas of the specification: Gravitational Fields and Space (the inverse-square gravitational field) and Further Mechanics (circular motion).

Explain why a geostationary satellite must orbit at one particular radius. By equating the gravitational force on the satellite to the centripetal force required to keep it in a circular orbit, derive Kepler's third law in the form T2=4π2GMr3.T^2 = \frac{4\pi^2}{GM}r^3.T2=GM4π2r3. Then calculate the radius of a geostationary orbit and the height of such an orbit above the Earth's surface.

Use the data: G=6.67×1011 N m2kg2G = 6.67\times10^{-11}\ \text{N m}^2\,\text{kg}^{-2}G=6.67×1011 N m2kg2; mass of the Earth ME=5.97×1024 kgM_E = 5.97\times10^{24}\ \text{kg}ME=5.97×1024 kg; orbital period T=24 h=86400 sT = 24\ \text{h} = 86400\ \text{s}T=24 h=86400 s; radius of the Earth RE=6.37×106 mR_E = 6.37\times10^{6}\ \text{m}RE=6.37×106 m.

(9 marks)

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Question 29 marksExplain

A photocell is used to investigate the photoelectric effect. Monochromatic light of two different frequencies is shone in turn on a clean metal photocathode. For each frequency the stopping potential VsV_sVs is found by making the collector electrode negative until the photocurrent just falls to zero. This experiment links Waves and the Particle Nature of Light (the photon model) with Electric Circuits (the role of the potential difference in stopping the electrons). The results are:

Frequency, fffStopping potential, VsV_sVs
5.00×1014 Hz5.00\times10^{14}\ \text{Hz}5.00×1014 Hz0.20 V0.20\ \text{V}0.20 V
9.00×1014 Hz9.00\times10^{14}\ \text{Hz}9.00×1014 Hz1.85 V1.85\ \text{V}1.85 V

Explain the photoelectric effect using the photon model - in particular why a threshold frequency exists, what is meant by the work function ϕ\phiϕ, why a stopping potential is needed, and why eVseV_seVs equals the maximum kinetic energy of the emitted electrons. Then use Einstein's photoelectric equation eVs=hfϕeV_s = hf - \phieVs=hfϕ at the two frequencies to determine Planck's constant hhh and the work function ϕ\phiϕ.

Use e=1.60×1019 Ce = 1.60\times10^{-19}\ \text{C}e=1.60×1019 C.

(9 marks)

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Question 36 marksCalculate

In a Core Practical, a student determines the specific heat capacity ccc of a metal block by the electrical method. An immersion heater is run for a measured time and the temperature rise is recorded, using c=VItmΔθ,c = \frac{VIt}{m\,\Delta\theta},c=mΔθVIt, where VVV and III are the heater pd and current, ttt is the heating time, mmm is the mass of the block, and Δθ=θ2θ1\Delta\theta = \theta_2 - \theta_1Δθ=θ2θ1 is the temperature rise. The measurements, each with its uncertainty, are:

QuantityValueUncertainty
Heater pd, VVV11.5 V±0.1\pm 0.1±0.1 V
Heater current, III4.0 A±0.1\pm 0.1±0.1 A
Heating time, ttt300 s±1\pm 1±1 s
Mass of block, mmm0.70 kg±0.001\pm 0.001±0.001 kg
Initial temperature, θ1\theta_1θ119.0 C19.0\ ^\circ\text{C}19.0 C±0.5 C\pm 0.5\ ^\circ\text{C}±0.5 C
Final temperature, θ2\theta_2θ241.0 C41.0\ ^\circ\text{C}41.0 C±0.5 C\pm 0.5\ ^\circ\text{C}±0.5 C

Calculate the specific heat capacity ccc. Then determine the percentage uncertainty in each quantity - taking care with the temperature rise Δθ\Delta\thetaΔθ - combine them correctly, and hence quote ccc with its absolute uncertainty.

(6 marks)

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Question 46 marksCalculate

In a Core Practical, a student determines the Young modulus EEE of the metal of a long, thin wire by hanging masses from it and measuring the extension. The Young modulus is given by E=FLAx,where the cross-sectional areaA=πd24.E = \frac{FL}{Ax}, \qquad \text{where the cross-sectional area} \quad A = \frac{\pi d^2}{4}.E=AxFL,where the cross-sectional areaA=4πd2. Here FFF is the stretching force, LLL is the original (unstretched) length, xxx is the extension produced, and ddd is the wire diameter, measured at several points with a micrometer and averaged. The measurements, each with its uncertainty, are:

QuantityValueUncertainty
Stretching force, FFF19.6 N±0.1\pm 0.1±0.1 N
Original length, LLL1.500 m±0.002\pm 0.002±0.002 m
Diameter, ddd0.40 mm±0.01\pm 0.01±0.01 mm
Extension, xxx1.8 mm±0.1\pm 0.1±0.1 mm

Calculate the cross-sectional area AAA and the Young modulus EEE. Then determine the percentage uncertainty in FFF, LLL, xxx and ddd, combine them correctly (noting that ddd is squared in the area), and hence quote EEE with its absolute uncertainty.

(6 marks)

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Question 56 marksEvaluate

In a Core Practical from Electric Circuits, a student determines the electromotive force ε\varepsilonε and internal resistance rrr of a cell. A variable load resistor is connected across the cell; for each setting the terminal pd VVV (voltmeter across the cell) and the current III (ammeter in series) are recorded. Because V=εIr,V = \varepsilon - Ir,V=εIr, a graph of VVV against III is a straight line whose intercept on the VVV-axis is ε\varepsilonε and whose gradient is r-rr.

The student takes a single pair of readings at each load setting and works fairly quickly through a wide range of currents.

Evaluate this experimental method. In your answer identify the independent, dependent and control variables; distinguish random from systematic error; explain why obtaining ε\varepsilonε from the intercept and rrr from the gradient of a best-fit line is more robust than using a single pair of readings; comment on reliability; and suggest improvements that would genuinely reduce the uncertainty (giving a reason for each).

(6 marks)

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Question 65 marksExplain

The same inverse-square (1/r21/r^21/r2) dependence appears in three different parts of the A-Level course:

  • the gravitational field strength a distance rrr from a point mass MMM, g=GMr2g = \dfrac{GM}{r^2}g=r2GM (Gravitational Fields and Space);
  • the electric field a distance rrr from a point charge QQQ, E=14πε0Qr2E = \dfrac{1}{4\pi\varepsilon_0}\dfrac{Q}{r^2}E=4πε01r2Q, and the closely related Coulomb force (Electric and Magnetic Fields);
  • the intensity of gamma radiation a distance rrr from a small point source, I1r2I \propto \dfrac{1}{r^2}Ir21 (Nuclear Radiation).

Explain why each of these is an inverse-square law, and state what the constant of proportionality is in each case.

(5 marks)

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