Oxford Revise AQA A Level Physics | Chapter P8 answers

P8: Materials

Question

Answers

Extra information

Mark

AO

Spec reference

01.1

\(
\begin{aligned}
& \text{Young Modulus = gradient}\\
& \frac{\left( 1200-600 \right) \times 10^6}{4.4-2.9} = \frac{600}{1.5}\\
& = 400\mathrm{\ MPa}\\
\end{aligned}
\)

Evidence of conversion of strain to decimal

Answer

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3.4.2.2

01.2

Below 250% strain, the stiffness is increasing; above 250%, it is constant

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3.4.2.2

01.3

\(
\begin{aligned}
& \text{Weight of car } = 1200\mathrm{\ kg\ } \times 9.8\text{ N kg}^{-1} = 11\ 760\ \mathrm{N}\\
& \text{Steel: yield stress is 250 MPa}\\
& \sigma = \frac{F}{A}\\
& A = \frac{F}{\sigma} = \frac{11\ 760}{250\ 000\ 000} = 7.2 \times 10^{-6}\mathrm{\ m}^{2}\\
& d = 2\sqrt{\frac{A}{\pi}} = 2\sqrt{\frac{7.2 \times 10^{-6}}{\pi} = 7.7 \times 10^{-3}}\mathrm{\ m}\\
& \text{Silk: yield stress is 1650 MPa}\\
& \sigma = \frac{F}{A}\\
& A = \frac{F}{\sigma} = \frac{11\ 760}{1\ 650\ 000\ 000} = 4.7 \times 10^{-5}\mathrm{\ m}^2\\
& d = 2\sqrt{\frac{A}{\pi}} = 2 \sqrt{\frac{4.7 \times 10^{-5}}{\pi} = 3.0 \times 10^{-3}}\mathrm{\ m}\\
\end{aligned}
\)

The diameter of the silk is less than half that of steel

Manipulation of equations

Answers

Comment

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3.4.2.2

01.4

\(
\begin{aligned}
& \text{Assume length of cable = 1 m}\\
& \mathrm{Weight } = mg, m = pV = p \pi r^2 = p \pi \ {\left( \frac{d}{2} \right)^2}\\
& \text{For steel: weight } = 7800 \pi \left( \frac{7.7 \times 10^{-3}}{2} \right)^2 \times 9.8 = 3.5\mathrm{\ N}\\
& \text{For silk: weight } = 1300 \pi \left( \frac{3.0 \times 10^{-3}}{2} \right)^2 \times 9.8 = 0.09\mathrm{\ N}\\
& \text{A steel cable has }\frac{3.5}{0.09} = 40\text{ times the weight of a silk cable}\\
\end{aligned}
\)

Use a length, or length cancels in ratio at the end

Correct use of weight and density equations

Ratio

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3.4.2.1

3.4.2.2

02.1

Using a vernier scale attached to the wire viewed through a microscope

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WS

02.2

250 N = limit of proportionality

275 N = yield point

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3.4.2.2

02.3

The cross-sectional area of the wire is decreasing/there is ‘necking’ of the wire

So the Young modulus is calculated from the initial section where the area is constant because the values of stress plotted used that area

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3.4.2.2

02.4

The material stretches beyond the yield point/shows plastic flow

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3.4.2.2

02.5

Line with twice the gradient

Line does not curve, stops abruptly

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3.4.2.2

03.1/

03.2/

03.4

03.1 Correct graph intersecting (40, 6)

Straight line through origin

03.2 Curved line up to force of 8 N

Unloading curve parallel to loading curve

03.4 Line of twice the gradient going through (20, 6)

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3.4.2.2

03.3

\(\mathrm{Energy } = \frac{1}{2}\ F \Delta l = \frac{1}{2} \times 6 \times 40 \times 10^{-3} = 0.12\mathrm{\ J}\)

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3.4.2.1

03.4

For springs in parallel, the same force will produce half the extension

\(\mathrm{Energy } = \frac{1}{2}\ F \Delta l = \frac{1}{2} \times 6 \times 20 \times 10^{-3} = 0.12\mathrm{\ J}\)

The energy stored is halved

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3.4.2.1

04.1

Measure the original length of the spring (when it is taut but not stretched)

Add known weight to the spring and find the new length, then subtract the original length to find the extension

Record length, force, and extension. Repeat and find the mean extension

Attach the weight to a higher point on the spring, and repeat to find the mean extension with the same weight

Repeat until the top of the spring is reached

Use at least 6 different lengths

Calculate the spring constants using \(k = \frac{F}{x}\) for each length

Plot a graph of spring constant (y-axis) against length (x-axis)

Measuring extension

Repeat measurements per length

Changing length to give sufficient range

Calculating k

Correct axes for graph

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3.4.2.1

04.2

Axes correctly labelled

Correct shape (inverse relationship, do not allow straight line with negative)

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3.4.2.1

04.3

As the length increases, the extension increases for the same force

\(k = \frac{F}{x}\), so the spring constant decreases

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3.4.2.1

04.4

\(\mathrm{strain } = \frac{\mathrm{extension}}{\mathrm{length}}\)

so gives the proportion by which the sample extends for a given force

Which is dependent on the material and not on the length of the sample.

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3.4.2.1

04.5

Example suggestion:

\(\mathrm{stress } = \frac{\mathrm{force}}{\mathrm{area}}\)

which could be modelled by having lots of springs in parallel

So the extension and spring constant depend on the number of springs for a given force, which is analogous to area

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3.4.2.1

05.1

\(
\begin{aligned}
\text{Area } & = \pi r^2\\
& = \pi \left( 0.11 \times 10^{-3} \right)^2 = 3.8 \times 10^{-8}\mathrm{\ m}^2\\
\end{aligned}
\) \(
\begin{aligned}
\sigma = \frac{F}{A} & = \frac{36\ \mathrm{N}}{3.8 \times 10^{-8}\ \mathrm{m}^2} & = 9.5 \times 10^{8}\ \mathrm{Pa}\\
\mathrm{Strain } \varepsilon & = \frac{\Delta l}{l_0} = \frac{0.66}{3.6} = 0.18\\
\end{aligned}
\) \(
\begin{aligned}
\textrm{Young modulus} & = \frac{\sigma}{\varepsilon}\\
& = \frac{9.5 \times 10^8\text{ Pa}}{0.18} = 5.2 \times 10^9\text{ Pa}\\
\end{aligned}
\)

Calculation of area

Calculation of stress

Calculation of strain

Answer

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3.4.2.2

05.2

\(
\begin{aligned}
& \text{Energy } = \frac{1}{2}\ F \Delta l\ = \frac{1}{2} \times 36\times 0.66 = 11.88\ \mathrm{J}\\
& \text{Mass of line } m = pV\\
& 1.15 \text{ g cm}^{-3} = 1150\ \text{kg m}^{-3}\\
& \mathrm{So},\ m = pV = 1150\ \text{kg m}^{-3} \times \left( {3.8 \times 10^{-8}\mathrm{\ m}^2 \times 3.6\ \mathrm{m}} \right) = 1.57 \times 10^{-4}\ \mathrm{kg}\\
\end{aligned}
\)

Assuming all of the energy stored is transferred to a kinetic energy store

\(
\begin{aligned}
& E = \frac{1}{2}\ mv^2\ \mathrm{OR}\ v = \sqrt{\frac{2E}{m}}\\
& = \sqrt{\frac{2 \times 11.88}{1.57 \times 10^{-4}\ \mathrm{kg}}}\\
& = 388\ \mathrm{m\ s}^{-1}\\
\end{aligned}
\)

Calculation of energy

Calculation of mass

Use of equation for kinetic energy

Answer

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3.4.2.2

3.4.1.8

05.3

If some energy is not transferred to the kinetic store, the speed would be smaller

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3.4.2.1

05.4

If the weight is halved, the energy is halved

Speed is proportional to \(\sqrt{E}\), so the speed is reduced by \(\frac{1}{\sqrt 2}\)

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3.4.2.2

3.4.1.8

06.1

Material A because there will be a small strain for a large stress

Do not award marks for just F and x without stress/strain

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3.4.1.1

06.2

Energy = area under graph by counting squares

1 square = 1 N × 0.005 m = 5 × 10–3 J, 42 squares (approximately)

0.21 J

\(\text{Allow area of triangle approximation } = \frac{1}{2} \times 0.05 \times 8 = 0.20\ \mathrm{J}\)

Allow 40–44 squares

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3.4.2.1

06.3

\(
\begin{aligned}
\text{Mass of cord } & = \text{ density } \times \text{ volume}\\
& = \text{density } \times \text{ length } \times \text{ area}\\
& = 1.15\text{ g cm}^{-3} \times 100\text{ cm} \times \pi \times (0.05\text{ cm})^{2}\\
& = 0.2875\text{ g}\\
& = 2.9 \times 10^{-4}\text{ kg}\\
\end{aligned}
\) \(\text{Energy } = \frac{1}{2}\ m{v^2}\) \(
\begin{aligned}
{c}v & = \sqrt{\frac{2E}{m}} \\
& = \sqrt{\frac{2 \times 4.2 \times 10^{-3}\ \mathrm{J}}{2.9 \times 10^{-4}\ \mathrm{kg}}}
& = 29(.3)\ \textrm{m s}^{-1}\\
\end{aligned}
\)

Use of density equation with consistent units

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3.4.2.1

3.4.1.8

06.4

Approximately half the extension for each force

Same shape

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3.4.2.2

06.5

\(
\begin{aligned}
\text{Initial gradient } = \frac{2\ \mathrm{N}}{0.008\ \mathrm{m}} & = 250\ \mathrm{N}\ \mathrm{m}^{-1}\\
\text{Young modulus } = \frac{\sigma}{\varepsilon} & = \frac{\left( \frac{F}{A} \right)}{ \left( \frac{x}{l} \right)} = \left( \frac{F}{x} \right) \left(\frac{l}{A}\right) \\
& = \frac{250}{\pi \left( 0.005 \right)^2}\\
& = 3.2(3.183) \times 10^6\ \text{N m}^{-2}\\
\end{aligned}
\)

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3.4.2.2

06.6

In equilibrium: RX + RY = mtotal g

RX + RY = (0.1 + 0.2 + 0.05) × 9.81 = 3.43 N

Clockwise moments about brick X: 0.1 × 9.81 × 0.45 + 0.2 × 9.81 × 0.55 + 0.05 × 9.81 × 0.5 = 0.441 + 1.08 + 0.245 =1.77 N m

Anticlockwise moment = 1 × RY

RY = 1.77 N

RX = 3.43 – 1.77 N = 1.67 N

Resolving forces vertically

Taking moments

Answers (both forces correct)

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3.4.1.2

07.1

The (average + ) diameter

The extension for each increase in load

Plot stress = \(\frac{\mathrm{load}}{\mathrm{area}}\) on the y-axis

Against strain = \(\frac{\mathrm{extension}}{\text{original length}}\) on the x-axis

Alternative

Plot load versus extension

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3.4.2.2

07.2

The strain will be too small/smaller than the actual value

Calculated Young modulus will be larger than value calculated with correct measurement

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3.4.2.2

07.3

Estimate of uncertainty = ± 2 mm

\(\% \text{ uncertainty } = \frac{{2 \times 10^{-3}\ \mathrm{m} \times 100}}{{60.0 \times 10^{-2}\ \mathrm{m}}} = 0.3\%\)

Accept values between 1 mm and 3 mm

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3.1.2

07.4

\(
\begin{aligned}
& \mathrm{Strain } = 0.1 \times 10^{-2}\\
& \mathrm{Stress } = \text{ Young modulus} \times \text{ strain } = 1.5 \times 10^{11} \mathrm{\ Pa\ } \times 10^{-3} = 1.5 \times 10^{8}\ \mathrm{Pa}\\
& = \frac{F}{A} = \frac{{1000\ {\mathrm{N}}}}{A}\\
& A = \frac{{1000\ {\mathrm{N}}}}{{1.5 \times {{10}^8}\ {\mathrm{Pa}}}} = 6.6 \times 10^{-6}\ m^{2}\\
& \text{Diameter}\ = 2 \times \sqrt{\frac{{6.6 \times {{10}^{-6}}\ {{\mathrm{m}}^{2}}}}{\pi }} = 2.9 \times {10^{-3}}\ {\mathrm{m}}\\
\end{aligned}
\)

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3.4.2.2

07.5

The force is shared by two wires

Maximum weight is 2000 N, so two people = 1400 N

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3.4.2.2

07.6

\(
\begin{aligned}
& {v^2} = {u^2} + 2as\\
& \text{Assuming acceleration is } 9.81 \text{ m s}^{-2}/\text{no air resistance}\\
& v = \sqrt{2as}\\
& v = \sqrt{2 \times 9.81 \times 12}\\
& = 15(.3)\text{ m s}^{-1}
\end{aligned}
\)

Correct assumption

Use of equation of motion

Answer to 2dp

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3.4.1.3

07.7

Lower

The drop of paint will reach terminal velocity faster than the brush because the weight is less

The brush will accelerate for a longer time reaching a larger speed

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3.4.1.4

08.1

Assume all the energy stored in the tendons is transferred to a gravitational potential energy store

\(
\begin{aligned}
& \frac{1}{2}\ F\Delta {\kern 1pt} l = mgh\\
& \mathrm{And } F = kx,\text{ so } mgh = \frac{1}{2}\ k{x^2}\\
& \mathrm{So } k = \frac{2mgh}{x^2}\\
& \text{Estimation of extension of tendon = 1 mm}\\
& \text{Assume height } = 10 \times 0.02 \mathrm{\ m\ } = 0.2 \mathrm{\ m}\\
& k = \frac{{2 \times 7 \times 10^{-6} \times 9.81 \times 0.2}}{{{{\left( 10^{-3} \right)}^2}}}\\
& = 27(.44)\ \text{N m}^{-1}\\
\end{aligned}
\)

Conservation of energy

Substitution for F

Estimation of extension (based on size of frog)

Calculation

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3.1.4.8

3.4.2.2

08.2

Straight-line graph labelled spring

Curved-line graph labelled rubber

Line labelled polythene

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3.4.2.1

08.3

Rubber bands

They are not permanently deformed when the load is removed

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3.4.2.1

08.4

\(
\begin{aligned}
& h = \frac{k{x^2}}{2mg}\\
& \text{The mass of the human is much bigger } \left( \frac{70\mathrm{\ kg}}{7\mathrm{\ g}} = 10^4 \right)\\
\end{aligned}
\)

so either the extension of the tendon would have to be 100 times bigger, or the tendons would need to be 10 000 times stiffer to produce the same height

Credit for any reason showing a link to the physical quantities used in part 04.1

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3.4.2.2

08.5

\(
\text{Energy stored } = mgh = \frac{1}{2}\ k{x^2},\text{ mass of human ~}70\mathrm{\ kg}
\) \(
\begin{aligned}
x & = \sqrt{\frac{2\ mgh}{k}} \\
x & = \sqrt{\frac{2 \times 70 \times 9.81 \times 1.5}{27}} \\
x & = 8.7\ {\mathrm{m}}
\end{aligned}\\
\)

The springs in the robot must have springs that are much stiffer than those of the tendon

Use of conservation of energy

Answer

Sensible comment

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3.1.4.8

3.4.2.2

08.6

\(
\begin{aligned}
\mathrm{Power } & = \frac{\text{energy}}{\text{time}}\\
& = \frac{mgh}{\text{time}}\\
& = \frac{70 \times 9.81 \times 1.5}{1.2} = 860\ \left( 858 \right)\ \mathrm{W}
\end{aligned}\\
\)

Which is about the same power as a microwave oven

Answer

Comment

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3.4.1.7

Skills box answers

Question

Answer

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\(
\begin{aligned}
& E = \frac{1}{A \times \text{ gradient}}\\
& E = \frac{2.82\ \mathrm{m}}{\pi \left( \frac{0.32 \times 10^{-3}}{2} \right)^2\ \mathrm{m}^2 \times 0.281 \times10^{-3}\ \mathrm{m\ N}^{-1}} = 1.25 \times {10^{11}}\ {\mathrm{Pa}}\\
& \text{% us in gradient } = \frac{\left( 0.315- 0.281 \right) \times 10^{-3}\ \mathrm{m\ N}^{-1}}{0.281 \times 10^{-3}\ \mathrm{m\ N}^{-1}} \times 100\% = 12.1\%\\
& \text{% us in diameter } = \frac{0.01\ \mathrm{mm}}{0.32\ \mathrm{mm}} \times 100\% = 3.1\%\\
& \text{% us in area } = 2 \times 3.1\% = 6.2\%\\
& \text{% us in length } = \frac{0.01\ \mathrm{mm}}{2.82\ \mathrm{mm}}\times 100\% = 0.35\%\\
& \mathrm{therefore } \% \text{ uc in }E = \left( 12.1 + 6.2 + 0.35 \right)\% = 18.7\%\\
& \text{uc in } E = 18.7\% \text{ of } 1.25 \times 10^{11} \mathrm{\ Pa\ } = \pm 0.23 \times 10^{11} \mathrm{\ Pa}\\
& \text{so } E = 1.3 \times 10^{11}\ \mathrm{Pa}\ \pm 0.2 \times 10^{11}\ \mathrm{Pa}\\
\end{aligned}
\)

2

The elastic limit has been exceeded. The wire has not returned to its original length on removing the load.

3

Rearranging the equation for the Young modulus to obtain the equation of a straight line \(\left( {y = mx + c} \right)\) gives\(F = \frac{{\left( {AE} \right)}}{l}e\).

The gradient is therefore equal to \(\frac{{AE}}{l}\).

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