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OCR 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 small bar magnet is suspended from the lower end of a vertical spring and set bobbing up and down. As it oscillates it passes in and out of a fixed flat coil that is connected to a data-logger, which records the induced EMF. This arrangement links two areas of the specification: Simple Harmonic Motion (from Circular Motion, SHM and Gravitational Fields) and Electromagnetic Induction (from Capacitors, Electric Fields and Electromagnetism).

The magnet of mass m=0.080 kgm = 0.080\ \text{kg}m=0.080 kg hangs on a spring of stiffness (force constant) k=12 N m1k = 12\ \text{N m}^{-1}k=12 N m1 and is released so that it oscillates with amplitude A=0.040 mA = 0.040\ \text{m}A=0.040 m.

Explain the motion and the signal recorded by the data-logger. Your answer should: show that the magnet performs simple harmonic motion (restoring force proportional to displacement) and use T=2πm/kT = 2\pi\sqrt{m/k}T=2πm/k to calculate the frequency fff and the maximum speed vmaxv_{\max}vmax; describe how the changing flux through the coil induces an EMF (Faraday's law) and what sets its direction (Lenz's law); and - the key synoptic point - explain where in the oscillation the induced EMF is greatest and where it is zero, justifying this from the speed of the magnet.

(9 marks)

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Question 29 marksExplain

A distant star can be modelled as a perfect black body. Astronomers measure the wavelength at which its emission is strongest and estimate its radius. This question links Astrophysics and Cosmology (Wien's displacement law and the Stefan-Boltzmann law) with Quantum Physics (the photon model).

For one such star the peak wavelength is λmax=5.0×107 m\lambda_{\max} = 5.0\times10^{-7}\ \text{m}λmax=5.0×107 m and the radius is r=7.0×108 mr = 7.0\times10^{8}\ \text{m}r=7.0×108 m.

Explain how the black-body model lets these two measurements be turned into other properties of the star, and carry out the steps. In your answer: use Wien's displacement law λmaxT=2.90×103 m K\lambda_{\max} T = 2.90\times10^{-3}\ \text{m K}λmaxT=2.90×103 m K to find the surface temperature TTT; use the Stefan-Boltzmann law L=4πr2σT4L = 4\pi r^2 \sigma T^4L=4πr2σT4 to find the luminosity LLL; and then use the photon model to estimate the rate at which the star emits photons, taking a typical photon to have the energy E=hcλmaxE = \dfrac{hc}{\lambda_{\max}}E=λmaxhc of a photon at the peak wavelength.

Use σ=5.67×108 W m2K4\sigma = 5.67\times10^{-8}\ \text{W m}^{-2}\,\text{K}^{-4}σ=5.67×108 W m2K4, h=6.63×1034 J sh = 6.63\times10^{-34}\ \text{J s}h=6.63×1034 J s and c=3.00×108 m s1c = 3.00\times10^{8}\ \text{m s}^{-1}c=3.00×108 m s1.

(9 marks)

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Question 36 marksDetermine

A student measures the speed of sound in air using a resonance tube - a tube closed at one end by an adjustable water level, with a vibrating tuning fork held over the open end. As the air column is lengthened, stationary waves form and the loudness passes through a series of maxima (resonances). The student records the first resonance length l1l_1l1 and the third resonance length l2l_2l2; for these two positions l2l1=λ2,l_2 - l_1 = \frac{\lambda}{2},l2l1=2λ, so the unknown end correction cancels. The tuning fork is stamped f=512 Hzf = 512\ \text{Hz}f=512 Hz (treat this as exact). The measured resonance lengths are:

QuantityValueUncertainty
First resonance, l1l_1l10.155 m±0.002\pm 0.002±0.002 m
Third resonance, l2l_2l20.490 m±0.002\pm 0.002±0.002 m

Determine the speed of sound vvv from these readings, and find the percentage uncertainty and hence the absolute uncertainty in vvv. Take care: because λ\lambdaλ comes from a difference of two lengths, the length uncertainties add.

(6 marks)

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Question 46 marksDetermine

A student determines the viscosity η\etaη of glycerol by timing a small steel ball bearing as it falls at terminal velocity through the liquid. At terminal velocity the weight is balanced by the upthrust and the viscous (Stokes') drag, which leads to η=2r2(ρsρf)g9vt,\eta = \frac{2r^2(\rho_s - \rho_f)g}{9\,v_t},η=9vt2r2(ρsρf)g, where rrr is the ball's radius, ρs\rho_sρs is the density of the steel, ρf\rho_fρf is the density of the fluid, g=9.81 m s2g = 9.81\ \text{m s}^{-2}g=9.81 m s2, and vtv_tvt is the terminal velocity. The measurements, each with its uncertainty, are:

QuantityValueUncertainty
Ball radius, rrr1.5×1031.5\times10^{-3}1.5×103 m±0.1×103\pm 0.1\times10^{-3}±0.1×103 m
Density of steel, ρs\rho_sρs7800 kg m⁻³±100\pm 100±100 kg m⁻³
Density of fluid, ρf\rho_fρf1260 kg m⁻³±20\pm 20±20 kg m⁻³
Terminal velocity, vtv_tvt0.045 m s⁻¹±0.002\pm 0.002±0.002 m s⁻¹

Determine the viscosity η\etaη, and find the percentage uncertainty and hence the absolute uncertainty in η\etaη. Take care: rrr is squared, so its percentage uncertainty must be doubled.

(6 marks)

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Question 56 marksEvaluate

In a Practical Activity Group (PAG) on simple harmonic motion, a student investigates how the period of a mass-spring system depends on the mass. A spring is clamped vertically, a mass mmm is hung from it, and the system is set oscillating. For each of several masses the student measures the period TTT. Because T=2πmkT2=4π2km  (+ a small intercept),T = 2\pi\sqrt{\frac{m}{k}} \quad\Rightarrow\quad T^2 = \frac{4\pi^2}{k}\,m \;(+ \text{ a small intercept}),T=2πkmT2=k4π2m(+ a small intercept), the student plots a graph of T2T^2T2 (on the yyy-axis) against mmm (on the xxx-axis), expecting a straight line.

The student times a single oscillation with a stopwatch for each mass, and starts timing from the highest point of the swing.

Evaluate this experimental method. In your answer identify the independent, dependent and control variables; distinguish random from systematic error; explain why timing many oscillations and plotting T2T^2T2 against mmm (using the gradient to find kkk, and noting what a non-zero intercept reveals) is more robust than a single timing; 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

A parabolic path turns up in two quite different parts of the course:

  • a projectile launched horizontally falls under gravity in a uniform gravitational field (Motion and Forces);
  • an electron fired horizontally between two oppositely charged, horizontal parallel plates is deflected in the uniform electric field between them (Capacitors, Electric Fields and Electromagnetism).

In each case the object enters horizontally with the same speed and follows a parabola while inside the field region.

Explain why both trajectories are parabolic. State the acceleration in each case (gravity: a=ga = ga=g; electron: a=eVmda = \dfrac{eV}{md}a=mdeV, where VVV is the plate pd and ddd their separation), and give one key difference between the two situations.

(5 marks)

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