Physics · AQA 7408 · Topic 3.6

Further mechanics and thermal physics

Circular motion, simple harmonic motion and resonance, then thermal physics: heating, ideal gases and kinetic theory.

  • 7 sub-topics
  • Taught in Year 13
  • 98 practice questions
  • 7 animated lessons
Practice questions (98) Lessons Worksheets All AQA Physics topics

What the specification says

Every sub-topic AQA can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. 3.6.1 Periodic motion
  2. 3.6.1.1Circular motionY13
  3. 3.6.1.2Simple harmonic motion (SHM)Y13
  4. 3.6.1.3Simple harmonic systemsY13
  5. 3.6.1.4Forced vibrations and resonanceY13
  6. 3.6.2 Thermal physics
  7. 3.6.2.1Thermal energy transferY13
  8. 3.6.2.2Ideal gasesY13
  9. 3.6.2.3Molecular kinetic theory modelY13

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 98 questions in this topic, free, no account needed.

Further mechanics

Not needed for Year 12Easy

Centripetal force · Year 13 · 1 mark

  1. An object of mass 800 kg moves at 6 m s⁻¹ in a circle of radius 0.5 m. What is the centripetal force on it?
    (1)

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  1. F=mv2rF = \dfrac{mv^2}{r}, directed towards the centre.C: 57600 N57600\ \mathrm{N}

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Oscillations

Not needed for Year 12Medium

A simple pendulum · Year 13 · 3 marks

A simple pendulum has length 0.90 m. Take g=9.81 m s−2g = 9.81\ \mathrm{m\,s^{-2}}.
  1. (a)
    Calculate its period.
    (2)
  2. (b)
    State why the swings must be small.
    (1)

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Part (a)2 marks

  1. For small swings a pendulum moves with simple harmonic motion.TT=2πlg{}= 2\pi\sqrt{\dfrac{l}{g}}=2π0.909.81{}= 2\pi\sqrt{\dfrac{0.90}{9.81}}
  2. Give 3 significant figures.T=1.90 sT = 1.90\ \mathrm{s}

Part (b)1 mark

  1. For large angles the motion is no longer simple harmonic.The formula uses sin⁡θ≈θ\sin\theta \approx \theta, which is only true for small angles.

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Oscillations

Not needed for Year 12Hard

Simple harmonic motion · Year 13 · 12 marks

Figure 1 shows how the displacement x of a 500 g mass on a spring varies with time t. The mass moves with simple harmonic motion.
t / sx / cm0.61.21.82.45−5
Figure 1
  1. (a)
    Use the graph to state the amplitude and the period of the motion.
    (2)
  2. (b)
    Calculate the angular frequency of the motion.
    (2)
  3. (c)
    Calculate the maximum speed of the mass.
    (2)
  4. (d)
    Calculate the maximum acceleration of the mass.
    (2)
  5. (e)
    Calculate the total energy of the oscillation.
    (2)
  6. (f)
    State a time, during the first oscillation, at which the kinetic energy of the mass is a maximum. Explain your answer.
    (2)

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Part (a)2 marks

  1. The greatest displacement from equilibrium.Amplitude = 5 cm
  2. The time for one complete oscillation.Period = 1.2 s

Part (b)2 marks

  1. Angular frequency from the period.ω\omega=2πT{}= \frac{2\pi}{T}=2π1.2{}= \frac{2\pi}{1.2}
  2. ω\omega=5.24 rad s−1{}= 5.24\ \mathrm{rad\,s^{-1}}

Part (c)2 marks

  1. Use the amplitude in metres.vmaxv_{max}=ωA{}= \omega A=5.24×0.05{}= 5.24 \times 0.05
  2. vmaxv_{max}=0.262 m s−1{}= 0.262\ \mathrm{m\,s^{-1}}

Part (d)2 marks

  1. The acceleration is greatest at maximum displacement.amaxa_{max}=ω2A{}= \omega^2 A=5.242×0.05{}= 5.24^2 \times 0.05
  2. amax=1.37 m s−2a_{max} = 1.37\ \mathrm{m\,s^{-2}}

Part (e)2 marks

  1. The total energy equals the kinetic energy at the equilibrium position.EE=12mvmax2{}= \tfrac{1}{2}m v_{max}^2=12×0.5×0.2622{}= \tfrac{1}{2} \times 0.5 \times 0.262^2
  2. E=0.0171 JE = 0.0171\ \mathrm{J}

Part (f)2 marks

  1. Read where the graph crosses the time axis.t = 0.3 s (or 0.9 s)
  2. The gradient of the graph is steepest there.The displacement is zero, so the mass is at the equilibrium position, where its speed is greatest.

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