NESA Physics Processing Data and Information

4 sample questions with marking guides and sample answers · Avg. score: 66.5%

Q5
2020
QCAA
Paper 2
4 marks
Q5
4 marks

An electron is situated halfway between two nuclei that are separated from each other by a distance of 4.5×10−104.5 \times 10^{-10} m. The first nucleus contains two protons. The second nucleus contains three protons.

Calculate the magnitude of the overall electromagnetic force experienced by the electron. (N to 1 decimal place)

Reveal Answer

F2=14πεoQ1×qer12F_2 = \dfrac{1}{4\pi\varepsilon_o}\dfrac{Q_1\times q_e}{r_1^2}

=9×109×3(1.6×10−19)×1.6×10−19(2.25×10−10)2= 9\times 10^9\times \dfrac{3(1.6\times 10^{-19})\times 1.6\times 10^{-19}}{(2.25\times 10^{-10})^2}

=1.3×10−8 N= 1.3\times 10^{-8}\ \text{N}

F1=14πεoQ2×qer22F_1 = \dfrac{1}{4\pi\varepsilon_o}\dfrac{Q_2\times q_e}{r_2^2}

=9×109×2(1.6×10−19)×1.6×10−19(2.25×10−10)2= 9\times 10^9\times \dfrac{2(1.6\times 10^{-19})\times 1.6\times 10^{-19}}{(2.25\times 10^{-10})^2}

=9.0×10−9 N= 9.0\times 10^{-9}\ \text{N}

∣Fnet∣=∣F1∣−∣F2∣|F_{net}| = |F_1| - |F_2|

=1.4×10−8−9.1×10−9= 1.4\times 10^{-8} - 9.1\times 10^{-9}

Force = 4.0×10−9 N4.0\times 10^{-9}\ \text{N} (to 1 decimal place)

Marking Criteria
DescriptorMarks

Indicates an understanding of the physical scenario in relation to Coulomb’s law.

1

Provides pertinent mathematical operation/s correctly performed.

1

Determines the forces (or electric field strength) imposed by each nuclei on the electron.

1

Determines the correct net force.

1
Q20
2024
VCAA
1 mark
Q20
1 mark

Data can be described as precise when

A

it is the result of a careful investigation.

B

the experiment is repeated many times, and the results show little variation.

C

the same experimental methodology is used by different investigators.

D

it is close to the scientifically accepted value of the quantity being measured.

Reveal Answer
A

it is the result of a careful investigation.

While a careful investigation is important in science, it does not define precision, which specifically refers to the consistency of repeated measurements.

B

the experiment is repeated many times, and the results show little variation.

Correct Answer

Precision refers to how close multiple measurements are to each other, meaning repeated trials will show very little variation.

C

the same experimental methodology is used by different investigators.

Having different investigators use the same methodology ensures standardization, but it does not guarantee that the resulting data will have low variance or high precision.

D

it is close to the scientifically accepted value of the quantity being measured.

Being close to the scientifically accepted or true value is the definition of accuracy, not precision.

Q1
2021
QCAA
Paper 2
3 marks
Q1
3 marks

A charge of 2.8×10−72.8 \times 10^{-7} C experiences an electrostatic force of 5.2×10−15.2 \times 10^{-1} N when placed near a charge of 3.2×10−73.2 \times 10^{-7} C.

Calculate the distance between the two charges. (m to 2 significant figures)

Reveal Answer

F=14πε0Qqr2F = \frac{1}{4\pi\varepsilon_0} \frac{Qq}{r^2}
r2=14πε0QqFr^2 = \frac{1}{4\pi\varepsilon_0} \frac{Qq}{F}
r=14πε0QqFr = \sqrt{\frac{1}{4\pi\varepsilon_0} \frac{Qq}{F}}
=9×109×2.8×10−7×3.2×10−70.52= \sqrt{9 \times 10^9 \times \frac{2.8 \times 10^{-7} \times 3.2 \times 10^{-7}}{0.52}}
Distance = 0.039 m (to 2 significant figures)

Marking Criteria
DescriptorMarks

Recognises the scenario relates to Coulomb’s law

1

Provides appropriate mathematical reasoning

1

Calculates the distance

1
Q15
2025
SCSA
13 marks
Q15

Electrons are accelerated from rest across a potential difference of 40.0 kV.

Q15a
5 marks

Calculate the final speed of the electrons using Newtonian physics, which ignores relativistic effects.

Reveal Answer

Ek=W=Vq=(1.60×10−19)(4.00×104)=6.40×10−15 JE_k = W = Vq = (1.60 \times 10^{-19})(4.00 \times 10^4) = 6.40 \times 10^{-15} \text{ J}

Ek=12mv2⇒v=2EkmE_k = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{\frac{2E_k}{m}}

Hence:
v=2(6.40×10−15)9.11×10−31=1.19×108 m s−1v = \sqrt{\frac{2(6.40 \times 10^{-15})}{9.11 \times 10^{-31}}} = 1.19 \times 10^8 \text{ m s}^{-1}

Marking Criteria
DescriptorMarks

Equates EkE_k to work done on electrons: Ek=W=VqE_k = W = Vq

1

Calculates value of EkE_k

1

Rearranges EkE_k formula for vv: Ek=12mv2⇒v=2EkmE_k = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{\frac{2E_k}{m}}

1

Substitutes values: v=2(6.40×10−15)9.11×10−31v = \sqrt{\frac{2(6.40 \times 10^{-15})}{9.11 \times 10^{-31}}}

1

Calculates answer: v=1.19×108 m s−1v = 1.19 \times 10^8 \text{ m s}^{-1}

1
Q15b
5 marks

Calculate the final speed of the electrons using Einstein’s special theory of relativity.

Reveal Answer

Et=Ek+Erest⇒Ek=Et−ErestE_t = E_k + E_{\text{rest}} \Rightarrow E_k = E_t - E_{\text{rest}}
Then:

Ek=mc21−v2c2−mc2=(1β−1)mc2 where β=1−v2c2\begin{align*} E_k &= \frac{mc^2}{\sqrt{1 - \frac{v^2}{c^2}}} - mc^2\\ &= \left(\frac{1}{\beta} - 1\right)mc^2 \text{ where } \beta = \sqrt{1 - \frac{v^2}{c^2}} \end{align*}

Solve for β\beta:

6.40×10−15=(1β−1)(9.11×10−31)(3.00×108)21β=6.40×10−15(9.11×10−31)(3.00×108)2+11β=1.0781β=0.9276\begin{align*} 6.40 \times 10^{-15} &= \left(\frac{1}{\beta} - 1\right)(9.11 \times 10^{-31})(3.00 \times 10^8)^2\\ \frac{1}{\beta} &= \frac{6.40 \times 10^{-15}}{(9.11 \times 10^{-31})(3.00 \times 10^8)^2} + 1\\ \frac{1}{\beta} &= 1.0781\\ \beta &= 0.9276 \end{align*}

Sub β\beta into β=1−v2c2\beta = \sqrt{1 - \frac{v^2}{c^2}}

0.9276=1−v2c2β=1−v2c2 and solves for v in terms of cv2c2=1−(0.9276)2v=0.1396 cv=0.3736 cv=1.12×108 m s−1\begin{align*} 0.9276 &= \sqrt{1 - \frac{v^2}{c^2}}\\ \beta &= \sqrt{1 - \frac{v^2}{c^2}} \text{ and solves for } v \text{ in terms of } c\\ \frac{v^2}{c^2} &= 1 - (0.9276)^2\\ v &= \sqrt{0.1396}\,c\\ v &= 0.3736\,c\\ v &= 1.12 \times 10^8 \text{ m s}^{-1} \end{align*}
Marking Criteria
DescriptorMarks

Uses mass-energy equivalence and total energy for EkE_k

1

Substitutes values

1

Solves for β\beta

1

Substitutes β\beta into β=1−v2c2\beta = \sqrt{1 - \frac{v^2}{c^2}} and solves for vv in terms of cc

1

Calculates answer

1
Q15c
3 marks

Calculate the percentage difference of your answer to part (a) compared to part (b).

Reveal Answer

1.19×108−1.12×1081.12×108×100=6.25%\frac{1.19 \times 10^8 - 1.12 \times 10^8}{1.12 \times 10^8} \times 100 = 6.25\%

Marking Criteria
DescriptorMarks

Correct numerator and correct denominator: 1.19×108−1.12×1081.12×108×100\frac{1.19 \times 10^8 - 1.12 \times 10^8}{1.12 \times 10^8} \times 100

2

Expresses as % difference: 6.25%6.25\%

1

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