A Level · Topics 12–25

A satellite is in a circular orbit at a height above the Earth's surface equal to the radius R of the Earth. The gravitational field strength at the Earth's surface is g. What is the acceleration of the satellite?

1. Ag
2. Bg/2
3. Cg/4
4. Dg/8

An ideal gas is at a thermodynamic temperature of 300 K. To what temperature must it be raised to double the root-mean-square speed of its molecules?

1. A424 K
2. B600 K
3. C1200 K
4. D2400 K

A body moves with simple harmonic motion of amplitude 2.0 cm and frequency 5.0 Hz. What is the magnitude of its maximum acceleration?

1. A0.63 m s⁻²
2. B3.1 m s⁻²
3. C20 m s⁻²
4. D200 m s⁻²

A capacitor of capacitance 470 μF, charged to a potential difference V₀, is discharged through a resistor of resistance 22 kΩ. After what time has the charge on the capacitor fallen to 25% of its initial value?

1. A3.6 s
2. B7.2 s
3. C14 s
4. D21 s

An electron enters a uniform magnetic field with its velocity perpendicular to the field. What is its subsequent path within the field?

1. Aa straight line at constant speed
2. Ba straight line with increasing speed
3. Ca circular arc at constant speed
4. Da parabolic arc with increasing speed

Light of wavelength 450 nm is incident on a metal surface of work function 2.0 eV. What is the maximum kinetic energy of the emitted photoelectrons?

1. A0.76 eV
2. B2.0 eV
3. C2.8 eV
4. D4.8 eV

A satellite is in a circular geostationary orbit about the Earth. The product GM for the Earth is 3.99 × 10¹⁴ m³ s⁻², and one day may be taken as 24 hours.

1. (a)

   Explain Explain what is meant by a geostationary orbit.

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   1. 1

      the satellite has a period of 24 hours, equal to the Earth's period of rotation

   2. 2

      it orbits in the same direction as the Earth's rotation, in the plane of the equator

   3. 3

      so it remains vertically above the same point on the Earth's surface

2. (b)

   Show (that) The gravitational force provides the centripetal force. Show that the radius of a geostationary orbit is approximately 4.2 × 10⁷ m.

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   1. 1

      GMm/r² = mrω², so GM = r³ω²

   2. 2

      ω = 2π/T = 2π/86400 = 7.27 × 10⁻⁵ rad s⁻¹

   3. 3

      r³ = GM/ω² = 3.99 × 10¹⁴ / (7.27 × 10⁻⁵)² = 7.55 × 10²²

   4. 4

      r = 4.2 × 10⁷ m

3. (c)

   Calculate Calculate the linear speed of the satellite in this orbit.

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   1. 1

      v = 2πr/T = 2π × 4.2 × 10⁷ / 86400

   2. 2

      v = 3.1 × 10³ m s⁻¹

4. (d)

   Explain A student suggests placing a geostationary satellite in orbit above London. Explain why this is not possible.

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   1. 1

      the centre of a circular orbit must be at the centre of the Earth, because that is where the gravitational force is directed

   2. 2

      an orbit whose plane passes above London at a fixed latitude would have its centre away from the Earth's centre

   3. 3

      so the satellite would have to be held up by some force other than gravity, which there is none to provide

A sealed cylinder contains an ideal gas of volume 2.0 × 10⁻³ m³ at a pressure of 1.5 × 10⁵ Pa and a temperature of 290 K. The Boltzmann constant is 1.38 × 10⁻²³ J K⁻¹.

1. (a)

   Calculate Calculate the number of molecules of gas in the cylinder.

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   1. 1

      Use the ideal-gas molecular relation pV = NkT for the sealed gas.

   2. 2

      N = pV/kT = (1.5 × 10⁵ × 2.0 × 10⁻³) / (1.38 × 10⁻²³ × 290)

   3. 3

      N = 7.5 × 10²²

2. (b)

   Determine Determine the total kinetic energy of the molecules in the cylinder.

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   1. 1

      mean kinetic energy per molecule = (3/2)kT = 1.5 × 1.38 × 10⁻²³ × 290 = 6.0 × 10⁻²¹ J

   2. 2

      total = N × 6.0 × 10⁻²¹

   3. 3

      total = 4.5 × 10² J (450 J)

3. (c)

   Deduce The gas is now compressed slowly at a constant temperature of 290 K. During the compression, 120 J of work is done on the gas. Deduce the change in internal energy of the gas and the thermal energy transferred to or from it.

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   1. 1

      the temperature is unchanged, so the mean kinetic energy per molecule is unchanged

   2. 2

      the internal energy of an ideal gas is entirely kinetic, so ΔU = 0

   3. 3

      first law: ΔU = q + w, so 0 = q + 120

   4. 4

      q = −120 J: 120 J of thermal energy is transferred out of the gas to the surroundings

4. (d)

   Explain Explain why the internal energy of an ideal gas depends only on its temperature, whereas that of a real gas does not.

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   1. 1

      an ideal gas is defined to have no forces between its molecules except during collisions, so there is no potential energy contribution and the internal energy is the total random kinetic energy alone

   2. 2

      a real gas has intermolecular forces, so its internal energy also includes potential energy, which depends on the separation of the molecules and hence on the volume

A mass suspended from a spring oscillates vertically with simple harmonic motion of amplitude 4.0 cm and period 0.80 s.

1. (a)

   Define Define simple harmonic motion.

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   1. 1

      motion in which the acceleration is proportional to the displacement from a fixed point

   2. 2

      and is always directed towards that point (opposite in direction to the displacement)

2. (b)

   Calculate Calculate the maximum speed of the mass.

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   1. 1

      ω = 2π/T = 2π/0.80 = 7.85 rad s⁻¹

   2. 2

      v_max = ωx₀ = 7.85 × 0.040

   3. 3

      v_max = 0.31 m s⁻¹

3. (c)

   Calculate Calculate the speed of the mass when its displacement from the equilibrium position is 2.0 cm.

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   1. 1

      v = ω√(x₀² − x²)

   2. 2

      v = 7.85 × √(0.040² − 0.020²) = 7.85 × 0.0346

   3. 3

      v = 0.27 m s⁻¹

4. (d)

   Explain The mass is now made to oscillate by a driver whose frequency can be varied, and the system is lightly damped. Describe how the amplitude of the oscillations varies with the driving frequency, and explain what happens to the response curve if the damping is increased.

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   1. 1

      the amplitude is small at low driving frequencies, rises to a sharp maximum when the driving frequency equals the natural frequency of the system, then falls again

   2. 2

      this maximum is resonance, and at it the driver transfers energy to the system most efficiently

   3. 3

      increasing the damping lowers the peak amplitude and makes the peak broader (flatter)

   4. 4

      and the peak occurs at a slightly lower frequency than the undamped natural frequency

Two large parallel metal plates are 5.0 mm apart in a vacuum. A potential difference of 2.0 kV is maintained between them. A separate circuit contains a capacitor of capacitance 22 μF charged to a potential difference of 12 V.

1. (a)

   Calculate Calculate the magnitude of the electric field strength between the parallel plates.

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   1. 1

      E = V/d = 2000 / 5.0 × 10⁻³

   2. 2

      E = 4.0 × 10⁵ V m⁻¹

2. (b)

   Explain Compare the field between the parallel plates with the field around an isolated point charge, and explain the difference in how each varies with position.

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   1. 1

      the field between the plates is uniform — the field lines are parallel and equally spaced, so E has the same magnitude and direction everywhere between them

   2. 2

      the field of a point charge is radial, with lines spreading out from the charge

   3. 3

      so the point-charge field falls off as 1/r², while the field between the plates does not vary with position at all

3. (c)

   Calculate Calculate the energy stored in the 22 μF capacitor.

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   1. 1

      W = ½CV² = ½ × 22 × 10⁻⁶ × 12²

   2. 2

      W = 1.6 × 10⁻³ J

4. (d)

   Determine The charged capacitor is discharged through a 150 kΩ resistor. Determine the time taken for the potential difference across it to fall to 3.0 V.

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   1. 1

      time constant RC = 150 × 10³ × 22 × 10⁻⁶ = 3.3 s

   2. 2

      V = V₀e^(−t/RC), so 3.0 = 12e^(−t/3.3)

   3. 3

      t = RC ln(V₀/V) = 3.3 × ln 4

   4. 4

      t = 4.6 s

Electrons are accelerated from rest through a potential difference of 250 V. Separately, a nucleus of ⁵⁶₂₆Fe has a nuclear mass of 55.9206 u. The mass of a proton is 1.00728 u, the mass of a neutron is 1.00867 u, and 1 u is equivalent to 931.5 MeV.

1. (a)

   Calculate Calculate the de Broglie wavelength of the accelerated electrons.

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   1. 1

      kinetic energy = eV = 1.60 × 10⁻¹⁹ × 250 = 4.0 × 10⁻¹⁷ J

   2. 2

      p = √(2mE_k) = √(2 × 9.11 × 10⁻³¹ × 4.0 × 10⁻¹⁷)

   3. 3

      p = 8.5 × 10⁻²⁴ kg m s⁻¹

   4. 4

      λ = h/p = 6.63 × 10⁻³⁴ / 8.5 × 10⁻²⁴ = 7.8 × 10⁻¹¹ m

2. (b)

   Explain Explain how the diffraction of a beam of such electrons by a thin crystal provides evidence for the wave nature of matter.

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   1. 1

      the beam produces a pattern of rings or maxima and minima rather than a single spot

   2. 2

      diffraction and interference are effects that only waves show

   3. 3

      the spacing of the pattern agrees with the de Broglie wavelength calculated from the electrons' momentum, so the wavelength is a property of the particles themselves

3. (c)

   Determine Determine the binding energy per nucleon of ⁵⁶₂₆Fe, in MeV.

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   1. 1

      total mass of separate nucleons = 26 × 1.00728 + 30 × 1.00867 = 56.4494 u

   2. 2

      mass defect Δm = 56.4494 − 55.9206 = 0.5288 u

   3. 3

      binding energy = 0.5288 × 931.5 = 492.6 MeV

   4. 4

      per nucleon = 492.6 / 56 = 8.8 MeV

4. (d)

   Explain Iron-56 lies near the peak of the binding energy per nucleon curve. Explain why energy is released both when light nuclei fuse and when heavy nuclei undergo fission.

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   1. 1

      energy is released whenever the products have a greater binding energy per nucleon than the reactants, because the nucleons end up more tightly bound and the surplus mass is released as energy

   2. 2

      light nuclei lie to the left of the peak, so fusing them moves the nucleons up the curve towards iron

   3. 3

      heavy nuclei lie to the right of the peak, so splitting them also moves the nucleons up the curve towards iron

A student investigates the discharge of a capacitor of capacitance C through a resistor of resistance R = 100 kΩ. The potential difference V across the capacitor is recorded at intervals, and ln(V/V) is calculated. The student expects V = V₀e^(−t/RC).

1. (a)

   Explain Explain why a graph of ln V against t should be a straight line if the discharge is exponential, and state what its gradient and intercept represent.

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   1. 1

      taking natural logarithms of V = V₀e^(−t/RC) gives ln V = ln V₀ − t/(RC)

   2. 2

      this is of the form y = mx + c with ln V as y and t as x, so the points lie on a straight line if the model holds

   3. 3

      the gradient is −1/(RC)

   4. 4

      the intercept on the ln V axis is ln V₀

2. (b)

   Determine Use the first and last data points to determine the gradient of the line, and hence determine the capacitance C.

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   1. 1

      gradient = (0.610 − 2.197) / (40.0 − 0.0) = −1.587 / 40.0

   2. 2

      gradient = −0.0397 s⁻¹

   3. 3

      RC = −1/gradient = 25.2 s

   4. 4

      C = 25.2 / 100 × 10³ = 2.5 × 10⁻⁴ F (250 μF)

3. (c)

   Determine The resistance is quoted as 100 ± 2 kΩ, and the uncertainty in the gradient is ±1.5%. Determine the percentage uncertainty in C, and state the value of C with its absolute uncertainty.

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   1. 1

      percentage uncertainty in R = (2/100) × 100 = 2%

   2. 2

      C is found by dividing by the gradient and by R, so the percentage uncertainties add: 1.5% + 2% = 3.5%

   3. 3

      absolute uncertainty = 0.035 × 250 = 8.75 ≈ 9 μF

   4. 4

      C = 250 ± 9 μF

4. (d)

   Suggest The voltmeter used has a finite resistance connected across the capacitor throughout the experiment. Suggest how this affects the value obtained for C, and suggest an improvement.

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   1. 1

      the voltmeter provides a second discharge path in parallel with R, so the effective resistance is less than 100 kΩ

   2. 2

      the capacitor therefore discharges faster than the model assumes; the measured time constant is too small, and since C is calculated using R = 100 kΩ, the value obtained for C is too small

   3. 3

      improvement: use a voltmeter (or digital multimeter) of much higher resistance, or a data logger with a high-impedance input