Oxford Revise A Level AQA Physics (Chapter 4)

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Here you’ll find all the answers to the activities and exam-style practice questions featured in Oxford Revise AQA A Level Geography.

Question

Answers

Extra information

Mark

AO

Spec reference

01.1

2.8 cm

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2

3.3.1.1

01.2

 \(
\begin{array}{lr}
\lambda = 2.8\ \textrm{cm}\ c = 8.4\ \rm{cm} \rm{s}^{-1}\\
c = f\lambda \\
f = \frac{c}{\lambda } = \frac{{8.4}}{{2.8}} = 3\,\,{\rm{Hz}}\\
\end{array}
\)

Allow e.c.f for λ answer from 01.1

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2

3.3.1.1

01.3

 \(\frac{\pi }{2}
\textrm{rad or} 90^{\circ}
\)

1 mark for phase difference,

1 mark for units

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1

2

1

3.3.1.1

01.4

 \(
\begin{array}{lr}
\textrm{Displacement will be negative (downwards) to max in}\frac{T}{4}\ \rm{s}\\
\textrm{Decreases through to zero displacement at} \frac{T}{2}\ \rm{s}
\end{array}
\)

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2

3.3.1.1

02.1

Double-ended arrow/line from between two points in phase

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1

3.3.1.1

02.2

Maximum displacement

From the equilibrium position (from rest position/mean position)

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1

3.3.1.1

02.3

Longitudinal wave

Spring oscillates in a direction parallel to/in energy transfer

OR because of compressions and rarefactions

Rarefaction must be spelt correctly for second mark

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1

2

3.3.1.2

02.4

Stationary wave formed

By superposition (or interference) (of two progressive waves)

Stationary points are nodes

Destructive interference where the spring is stationary

Max of 3 marks awarded

max 3

2

3.3.1.3

03.1

With unpolarised light, the vibrations are in many planes

In plane polarised light, the oscillations are in one plane only

ignore ‘direction’

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1

3.3.1.2

03.2

Reflected light is polarised

So intensity of light reflected on water will be reduced by polarising filter

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1

2

3.3.1.2

03.3

Rotate the polarising filter through 180°/360°

Variation in intensity between max and min (or light and dark)

One maxima and min in 180° OR two maxima (or two minima) in 360° rotation

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3

3.3.1.2

03.4

Sound waves are longitudinal waves

Since oscillations are parallel to/same direction as wave travel, they cannot be polarised

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3.3.1.2

03.5

Sketch:

  • showing ray of light reflected away from the normal as it leaves the water
  • straight line drawn showing where ray appears to come from person’s perspective

Explanation that change of speed at boundary causes refraction

Max 2 marks for two rays drawn and 1 mark for explanation

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2

3.3.2.3

04.1

Reflection from metal plate

Two waves of the same frequency/wavelength

travelling in opposite directions (or forward/reflected waves)

Maxima where waves are in phase or interfere constructively

Minima where waves are out of phase/antiphase or interfere destructively

Nodes and antinodes or stationary waves identified

Any three awarded

max 3

2

3.3.1.3

04.2

 \(
\begin{array}{lr}
\textrm{Distance between minima is}\ \frac{\lambda }{2}\\
4 \times \frac{\lambda }{2}= 54\ \rm{mm}\\
\lambda = 27\ \rm{mm}
\end{array}
\)

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1

2

3.3.1.3

3.3.1.1

04.3

\(
\begin{array}{lr}
c =f\ \lambda\ \textrm{and}\ c = 3.0 \times 10^8\ \rm{m} \rm{s}^{-1}\\
f = \frac{c}{\lambda } = \frac{{3.0 \times {{10}^8}}}{{27 \times {{10}^{ – 3}}}} = 1.1 \times 10^{10}\ \rm{Hz}\\
11\ \rm{GHz}
\end{array}
\)

Allow e.c.f from 04.2

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2

3.3.1.3

3.3.1.1

3.1.1

04.4

P labelled close to the plate in direct line with transmitter

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1

3.3.1.3

04.5

The distance travelled by the transmitted wave and the reflected wave is similar at point P

The amplitude of both waves will be similar

Max destructive interference

Max of two points

max 2

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3.3.1.3

04.6

The microwave transmitter produces plane polarised waves and so the detector must be in the correct plane

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3

3.3.1.2

05.1

Stationary waves set up in the microwave

Microwaves are reflected in the oven

Two waves of the same frequency/wavelength travelling in opposite directions (or forward/reflected waves)

Melted marshmallows are where waves are in phase or interfere constructively

The melted marshmallows are at an antinode

Any three from

max 3

2

3.3.1.3

05.2

 \(
\begin{array}{lr}
c = f\ \lambda\ \textrm{and}\ c = 3.0 \times 10^8\ \textrm{m s}^{-1}\\
\lambda = \frac{c}{f} = \frac{{3.0 \times {{10}^8}}}{{2450 \times {{10}^6}}} = 0.122\ \rm{m}\\
\textrm{Distance between minima is}\ \frac{\lambda }{2}\\
= 0.061\ \rm{m}
\end{array}
\)

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1

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2

3.3.1.1

3.3.1.3

3.1.1

05.3

 \(
\begin{array}{lr}
c = f \lambda = 2450 \times 10^6 \times 0.142\ \rm{m} = 3.48 \times 10^8\ \textrm{m s}^{-1}\\
\% \,{\rm{error}} = \frac{{{\rm{difference}}}}{{{\rm{actual}}}} \times 100\%\\
\% \,{\rm{error}} = \frac{{{4}{\rm{.8}} \times {1}{{0}^7}}}{{{3}{\rm{.0}} \times {1}{{0}^8}}}
\times 100\% = 16\%
\end{array}
\)

Allow e.c.f from value of c

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2

3.3.1.3

3.1.2

05.4

They would have to remove/disable the turntable so that marshmallows were stationary

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3

05.5

In a progressive wave, all particles vibrate at same frequency

AND in a stationary wave, all particles except those at nodes vibrate at same frequency

In a progressive wave, all particles have same amplitude

AND in stationary waves, particles vary in amplitude from zero at nodes to max at antinodes

\(
\textrm{In progressive waves, the phase difference between particles} =
\frac{{2\pi d}}{\lambda },\ \textrm{where}\ d \ \textrm{is the distance apart}
\)

AND in stationary waves, phase difference = nπ, where n is the number of nodes between the particles

Each statement must compare stationary and progressive waves

2 marks maximum for each property compared

Two properties compared for maximum marks

max 4

2

3.3.1

06.1

Waves travel to the boundaries and are reflected

Two waves travelling in opposite directions interfere/superpose

Antinodes or maxima where waves are in phase or interfere constructively

OR

Nodes/minima where waves are out of phase/antiphase or interfere destructively

Nodes form at fixed ends

NOT bounces off

max 3

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3.3.1.3

06.2

λ = 0.62 × 2 = 1.24 m

c = f λ = 82.41 × 1.24 = 102 m s−1

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3.3.1.1

06.3

 \(f = \frac{c}{\lambda } = \frac{102}{{0.62}}= 165\ \rm{Hz}\)

Allow e.c.f from 06.2

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3.3.1.1

06.4

 \(
\begin{array}{lr}
\textrm{Use of}\ f = \frac{1}{{2l}}\sqrt {\frac{T}{\mu }} \\
f{^2}\, \propto \frac{1}{\mu }\\
{f^{2}}\,\mu = \textrm{constant} = 82.41^{2} \times 6.44 \times 10^{-3}\\
\rm{New}\ \mu = \frac{{{{82.41}^2} \times 6.44 \times {{10}^{ – 3}}}}{{{{196}^2}}} = 1.14 \times 10^{-3}\ \rm{kg}\ \rm{m}^{-1}
\end{array}
\)

Alternatively, they could work out T from initial info and calculate

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3

3.3.1.3

07.1

3rd/third harmonic

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1

3.3.1.3

07.2

P antinode AND constructive interference/in phase

Q node AND destructive interference/out of phase

Need name and description for each mark

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3.3.1.3

07.3

P has positive amplitude and Q negative

P and Q are π rad or 180° out of phase

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3.3.1.3

3.3.1.1

07.4

 \(
\begin{array}{lr}
\lambda = \frac{2}{3}L\ = 0.80\ \rm{m}\\
c = f \lambda\\
f = \frac{c}{\lambda } = \frac{{13.6}}{{0.80}} = 17\ \rm{Hz}
\end{array}
\)

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2

3.3.1.1

3.3.1.3

07.5

6th harmonic sketched – expect to see 6 antinodes and 7 nodes

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3.3.1.3

08.1

 \(
\begin{array}{lr}
80\ \textrm{(ms)}\\
f = \frac{1}{T}= 12.5 Hz\\
f^2 = 156\ \rm{Hz}
\end{array}
\)

Answer in table row should be completed

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2

3.3.1.1

08.2

\(
\begin{array}{lr}
T = mg\\
f = \frac{1}{{2l}}\sqrt {\frac{T}{\mu }} =\frac{1}{{2l}}\sqrt {\frac{{mg}}{\mu }}\\
f{^2}\, = \frac{1}{{4{l^2}}} \times \frac{{mg}}{\mu }\\
\textrm{Since}\ l,\ g,\ \textrm{and}\ \mu are constant\\
{f^2} \propto m
\end{array}
\)

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1

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3.3.1.3

08.3

 \(\textrm{Mark for plotting point} \pm \frac{1}{2} \textrm{square on graph}\)

Mark for drawing line of best fit

Large triangle drawn or evidence shown

1.2 ± 0.1 (g Hz−2 )

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MS 3.2, 3.3, 3.4

08.4

 \(
\begin{array}{lr}
\textrm{Gradient} =\frac{g}{{4{l^2}\mu }}\\
\mu = \frac{{9.81}}{{4 \times {1^{{\kern 1pt} 2}} \times 1.2}}\\
2.0\ \rm{g}\ \rm{m}^{-1}
\end{array}
\)

e.c.f

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2

3.3.1.3

08.5

 \(
\begin{array}{lr}
\textrm{% uncertainty in length} =
\frac{{0.001}}{{1.000}} \times 100\% = 0.1\%\\
\textrm{OR % uncertainty in mass}=\frac{{0.1}}{{1.7}} \times 100\% = 5.9\%\\
\textrm{total % error = 6.0%}\\
\textrm{absolute error} = 0.06 \times 1.7 = \pm\ 0.1\ \rm{g}\ \rm{m}^{-1}
\end{array}
[//latex]

1 mark for calculating either % uncertainty

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3.1.2

08.6

[latex]
\begin{array}{lr}
\% \,\,{\rm{difference}} = \frac{{{\rm{difference}}}}{{{\rm{actual}}}}\times 100\%\\
\% \,\,{\rm{difference}} = \frac{{{0}{\rm{.3}}}}{{{1}{\rm{.7}}}} \times\ 100\% = 18\%
\end{array}
\)

Possible e.c.f from their value for 08.4

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3.1.2

Skills box answers

Question

Answer

1

Wear goggles in case the wire snaps. Place a box containing padding material below the masses in case they fall.

2

 \(
\textrm{Set up the apparatus as shown in the diagram. Record values of}\ f\ \textrm{for different values of}\ l.\ \textrm{Change the length in the string by moving the vibrator closer to the pulley.}\
f = \frac{1}{{2l}}\sqrt {\frac{T}{\mu }} ,\ \textrm{therefore plot a graph of}\ l\ \textrm{against}\ \sqrt{T},\ \textrm{will produce a straight line through the origin, confirming the relationship between}\ l\ \textrm{and}\ T.
\)

3

 \(
\begin{array}{lr}
{\rm{The\;gradient\;of\;the\;graph}} = \frac{1}{{\left( {4{l^2}\mu } \right)}}\\
\textrm{so} = (4l^2 gradient)^{-1}
\end{array}
\)
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