Oxford Revise AQA A Level Physics | Chapter P1 answers

P1: Measurements and errors

Question

Answers

Extra information

Marks

AO

Spec reference

01.1

v = 2.54

v2 = 6.43 (If rounded data used, may have 6.45)

Must be to same number of sig figs as table

2

2

3.4.1.3

01.2

Point plotted to within nearest half grid square

Line of best fit drawn – with intercept

1

1

2

MS 3.2

01.3

Systematic

Error in measuring height
s – needs to be middle of card to middle of light gate/s; measured too short

1

1

3

3.1.2

01.4

Triangle drawn on graph or use of coordinates demonstrated

Value for gradient = 17 ± 0.4

Not simply using values from table – must be from graph

1

1

2

MS 3.4

01.5

\(
\begin{aligned}
& v^2 = u^2 + 2as \text{ and } u = 0\\
& v^2 = 2as\\
& \text{Gradient } = 2a \text{ or } 2g\\
& g = \frac{17}{2} = 8.5\ (\mathrm{m\ s}^{-2})
\end{aligned}
\)

Range of values 8.7 to 8.3 from gradient

1

1

1

2

3.3.1.3

01.6

\(
\begin{aligned}
& \%\text{ difference } = \frac{\text{result} – \text{actual}}{\text{actual}} \times 100\%\\
& \%\text{ difference } = \frac{1.31}{9.81} \times 100\% = 13\%
\end{aligned}
\)

ignore minus sign

1

2

3.1.2

02.1

Max three from:

  • draw round the semi-circular block and mark a point in the centre of the straight edge (measured with ruler)
  • use a protractor to mark the normal (line perpendicular) from this point
  • dark lines 5° apart from the normal to 35° (at least 6 suggested) – each entering the glass at 45°
  • use fine ray of light or laser as source
  • point along the drawn pathways mark a point to trace the path of the outgoing ray.

max 3

2

Atj

02.2

Large triangle seen or coordinates used shown \(\frac{0.7-0.06}{0.5-0}=1.3\)

triangle is at least half the graph

Accept 1.28

1

1

2

3.3.2.3

02.3

\(
\begin{aligned}
& \text{Line of worst fit drawn (could be max or min)}\\
& \text{Gradient of max }=\frac{0.9-0.0}{0.64-0.02}=1.5\\
& \text{Absolute uncertainty }=1.5-1.3=0.2\\
& \text{Gradient of min }=\frac{0.88-0.1}{0.66-0.00}=1.2\\
& 1.3 \times 1.2=0.1
\end{aligned}
\)

Full 3 marks for steepest as that is line of worst fit

Min line gains 2 marks

1

1

1

2

3.1.2

02.4

Experimental value = 1.3 ± 0.2, therefore value does lie within experimental uncertainty

Answer consistent with their results – so if they drew min line of best fit, value does not lie in experimental uncertainty

1

1

3.1.2

03.1

Temperature:

  • place whole apparatus in water bath making sure trapped air completely submerged
  • stir regularly to make sure temperature even
  • leave time at each temperature to ensure air at same temperature as water bath
  • use thermometer/digital thermometer to record temperature.

Volume:

  • attach apparatus to a ruler
  • read length of trapped air – make sure eye level with meniscus when reading
  • do not remove from water bath when reading measurement.

Must have at least one statement from volume and one from temperature for full marks

max 4

2

3.1.2

03.2

Max two from:

  • lowest temperature possible
  • minimum internal energy (allow zero kinetic energy)
  • −273 °C
  • pressure of a gas at this temperature is zero.

max 2

1

3.6.2.2

03.3

\(
\begin{aligned}
& \text{Intercept } = 3\\
& \text{Gradient } = \frac{3.9 – 3.0}{80} = 0.011\\
& \text{Use of } y = mx + c \text{ when } y = 0\\
& 0 = 0.011x + 3\\
& x = -270\ (272)\ (^{\circ}\mathrm{C})
\end{aligned}
\)

1

1

1

1

2

M3.3 and 3.4

3.6.2.2

03.4

This value is much lower than earlier value

This will be because the air was warmer than the surrounding water as the water cooled too quickly/temperature lag

This would give a larger intercept and shallower gradient making result too low

1

1

1

3

3.1.2

04.1

Using a micrometer/Vernier callipers

Take the diameter along the length in several places and find mean

Idea of several readings along length important for second mark

1

1

2

3.1.2

04.2

\(
\begin{aligned}
& A = \pi r^2 = \pi \times (0.11 \times 10^{-3})^2 = 3.8 \times 10^{-8}\ \mathrm{m}^2\\
& \text{Use of } \rho = \frac{RA}{l} = \frac{7.0 \Omega \times 3.8 \times 10^{-8}}{0.50} = 5.3 \times 10^{-7}\\
& \Omega \mathrm{\ m}
\end{aligned}
\)

1

1

1

2

3.5.1.3

04.3

\(
\begin{aligned}
& \text{% length } = \frac{0.001}{0.500} \times 100\% = 0.2\%\\
& \text{% diameter } = \frac{0.01}{0.22} \times 100\% = 4.5\%\\
& \%\ R = \frac{0.4}{7.0} \times 100\% = 5.7\%\\
& \text{% uncertainty } = 0.2 + (2 \times 4.5) + 5.7 = 15\%
\end{aligned}
\)

1 mark for calculating any one percentage uncertainty correctly

1

1

2

3.1.2

04.4

Any sensible suggestion for graph and how
ρ calculated:

  • plot R versus length, and gradient = \(\frac{\rho}{A}\)
  • plot
    RA versus length, and gradient =
    ρ

Why more accurate:

  • allows you to identify anomalies
  • systematic errors in measuring length/or resistance of connecting wires will not affect final answer.

Must have explained graph and suggested why more accurate for full marks

max 3

3

3.5.1.3

3.1.2

05.1

Depth:

  • using a ruler with no zero error/marking levels on side of tray before experiment begins
  • ensure ruler is read at eye level.

Speed:

  • using a stopwatch
  • measure the time for wave to travel at least three lengths of tray/or some consideration of increased distance/increased time to reduce uncertainty caused by human reaction time.

Allow any acceptable method for accurate measurements

1

1

1

1

3

PS1.2

05.2

Any sensible suggestion of graph, what to expect and how to confirm:

v versus \(\sqrt h\)

should be straight-line graph through the origin

gradient = \(\sqrt g\)

or

v2 versus
h

should be straight-line graph through the origin

gradient = g

1

1

1

3

3.1.2

MS3.3

05.3

Students would have to confirm by using a different set of apparatus;

or see if another student found the same relationship repeating the experiment

1

1

3.1.2

06.1

Sensible guess at room’s dimensions, such as:

3 m × 10 m × 15 m = 450 m3

Use of mass =
ρV

mass = 1.2 × 450 = 540 kg

Allow any sensible proposal here

1

1

3

3.1.3

06.2

80 × 28 = 2240 J s−1

1

2

3.4.1.7

06.3

\(
\begin{aligned}
& E = 2240 \times 20 \times 60 = 268\ 8000\ \mathrm{J}\\
& E = m\ c\ \Delta\theta\\
& \Delta \theta = \frac{E}{mc} = 5\ ^{\circ}\mathrm{C}
\end{aligned}
\)

Answers will vary based on mass calculations

1

1

2

3.6.2.1

06.4

No

As temperature in room rises, thermal energy will be transferred from the room

Or rate of energy transfer dependent on temperature outside/temperature difference/insulation/of walls/windows doors

Sensible answer backed by logical reasoning

1

1

3

3.6.2.1

07.1

Labelled diagram of apparatus – may be marks available here for method depending on detail

  • method for small amplitudes – for example, protractor (less than 10°) or method of measuring start height accurately each time
  • timing multiple oscillations to reduce uncertainty from reaction time using stopwatch
  • fiducial marker at centre point for timing
  • measuring length from top to middle of bob with metre ruler
  • lengths chosen so
    T longer
  • evidence that clamp stand is clamped to the desk for safety – or bag under bob in case it falls.

1 mark for diagram, 3 further marks for detail on method and accuracy

max 4

2

Atb and d

PS2.1

PS4.1

07.2

In first column 2 becomes 2.00

Add units s2 for third column

Values correct in third column:

8.12, 7.67, 7.40, 7.13, 6.60

1

1

1

2

PS2.2

07.3

Points plotted correctly within ½ square

Line of best fit drawn

Gradient calculated = 4.0 ± 0.2

1

1

1

2

MS3.2

07.4

\(
\begin{aligned}
& T = 2\pi \sqrt {\frac{l}{g}} \text{ so } {T^2} = 4{\pi ^2}\frac{l}{g}\\
& \text{Gradient } = \frac{4\pi^2}{g}\\
& g = 9.9\ \mathrm{m\ s}^{-2}
\end{aligned}
\)

1

1

2

3.6.1.3

07.5

The data is precise because the points are close to the line of best fit

The data is accurate because the value of
g is within 1% of the actual value

For second mark there should be justification of accuracy

1

1

1

3.1.2

08.1

\(
\begin{aligned}
& \text{Use of } Q = It\text{ or } W = VQ\\
& W = Fd = mad\ (\text{or other energy equation})\\
& V = \frac{W}{Q} = \frac{W}{It}\\
& V = \frac{mad}{it} = \frac{\mathrm{kg\ m\ s}^{-2} \mathrm{\ m}}{\mathrm{A\ s}} = \mathrm{kg\ m}^{2}\ \mathrm{A}^{-1}\ \mathrm{s}^{-3}
\end{aligned}
\)

Must be able to see cancelling and evidence of equations (can be entirely in units)

1

1

1

3

3.1.1

08.2

Simple circuit with one cell, variable resistor and ammeter in series, and voltmeter in parallel with variable resistor or cell

Mark for correct symbols and mark for correct arrangement

1

1

1

3.5.1.6

08.3

E.m.f. is
y-intercept – 1.51 ± 0.05 V

Internal resistance is the gradient 0.41 ± 0.2 Ω

1

1

2

3.5.1.6

08.4

The actual e.m.f. is lower – or stated value 1.41 – consistent with results

Internal resistance will be the same because all points are shifted by same amount

For all 3 marks students must have explained answers

1

1

1

3

3.1.2

Skills box answers

Question

Answer

1(a)

\(\frac{0.1}{6.7}\times 100\% = 1.49\%\)

1(b)

\(\frac{10}{450} \times 100\% = 2.22\%\)

1(c)

\(\frac{1000}{366\ 000} \times 100\% = 0.27\%\)

2(a)

3.43 × 6.5% = 3.43 × 0.065 = 0.22

= 3.43 W ± 0.22 W

2(b)

10 × 10% = 10 × 0.1 = 1

= 10 kΩ ± 1 kΩ

2(c)

12 742 × 0.3% = 12 742 × 0.003 = 38.2

= 12 742
km ± 38.2
km

3(a)

volume of cube = \(\left( 25.0 \times 10^{-3}\mathrm{\ m} \right)^3 = 1.56 \times 10^{-5}\mathrm{ m}^3\)

Calculate the percentage uncertainty in each measurement first:

\(\left( \frac{0.2\ \times 10^{-3}} {25.0 \times 10^{-3}} \right) \times 100\% = 0.8\%.\)

Then add the percentage uncertainties for all three measurements, which is 2.4%.

So absolute uncertainity \(= 2.4\ \% \text{ of } 1.56 \times 10^{-5}\ \mathrm{m}^3 = 4 \times 10^{-7}\ \mathrm{m}^3\).

Note: standard form integers must be between ≥ 1 or < 10, so the power of the absolute uncertainty value has changed in this answer. Remember to always check powers when using standard form.

3(b)

\(\text{density} = \frac{\text{mass}}{\mathrm{volume}} = \frac{42.19 \times 10^{-3}\ \mathrm{kg}}{1.56 \times 10^{-5}\ \mathrm{m}} = 2700\ \mathrm{kg\ m}^{-3}\)
.

To calculate uncertainty in density you need to add the percentage uncertainties in volume and mass.

\(\text{% uc mass } = \left( \frac{0.01}{42.19} \right) \times 100\% = 0.024 \%\)

Adding the % uc = (0.024 + 2.4)% = 2.4%

Absolute uncertainty = absolute value × % uc = 64.8 kg m−3

The density of aluminium = 2700 kg m−3 ± 65 kg m−3.

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