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NESA Physics Processing Data and Information

3 sample questions with marking guides and sample answers · Avg. score: 65.9%

2021

QCAA

Paper 2

3 marks

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

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

$F = \frac{1}{4\pi\varepsilon_0} \frac{Qq}{r^2}$
$r^2 = \frac{1}{4\pi\varepsilon_0} \frac{Qq}{F}$
$r = \sqrt{\frac{1}{4\pi\varepsilon_0} \frac{Qq}{F}}$
$= \sqrt{9 \times 10^9 \times \frac{2.8 \times 10^{-7} \times 3.2 \times 10^{-7}}{0.52}}$
Distance = 0.039 m (to 2 significant figures)

Marking Criteria

Descriptor	Marks
Recognises the scenario relates to Coulomb’s law	1
Provides appropriate mathematical reasoning	1
Calculates the distance	1

2020

QCAA

Paper 2

4 marks

An electron is situated halfway between two nuclei that are separated from each other by a distance of $4.5 \times 10^{-10}$ 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)

$F_2 = \dfrac{1}{4\pi\varepsilon_o}\dfrac{Q_1\times q_e}{r_1^2}$

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

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

$F_1 = \dfrac{1}{4\pi\varepsilon_o}\dfrac{Q_2\times q_e}{r_2^2}$

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

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

$|F_{net}| = |F_1| - |F_2|$

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

Force = $4.0\times 10^{-9}\ \text{N}$ (to 1 decimal place)

Marking Criteria

Descriptor	Marks
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

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.

$E_k = W = Vq = (1.60 \times 10^{-19})(4.00 \times 10^4) = 6.40 \times 10^{-15} \text{ J}$

$E_k = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{\frac{2E_k}{m}}$

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

Marking Criteria

Descriptor	Marks
Equates $E_k$ to work done on electrons: $E_k = W = Vq$	1
Calculates value of $E_k$	1
Rearranges $E_k$ formula for $v$ : $E_k = \frac{1}{2}mv^2 \Rightarrow v = \sqrt{\frac{2E_k}{m}}$	1
Substitutes values: $v = \sqrt{\frac{2(6.40 \times 10^{-15})}{9.11 \times 10^{-31}}}$	1
Calculates answer: $v = 1.19 \times 10^8 \text{ m s}^{-1}$	1

Q15b

5 marks

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

$E_t = E_k + E_{\text{rest}} \Rightarrow E_k = E_t - E_{\text{rest}}$
Then:

\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 $\beta$ :

\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 $\beta$ into $\beta = \sqrt{1 - \frac{v^2}{c^2}}$

\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

Descriptor	Marks
Uses mass-energy equivalence and total energy for $E_k$	1
Substitutes values	1
Solves for $\beta$	1
Substitutes $\beta$ into $\beta = \sqrt{1 - \frac{v^2}{c^2}}$ and solves for $v$ in terms of $c$	1
Calculates answer	1

Q15c

3 marks

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

$\frac{1.19 \times 10^8 - 1.12 \times 10^8}{1.12 \times 10^8} \times 100 = 6.25\%$

Marking Criteria

Descriptor	Marks
Correct numerator and correct denominator: $\frac{1.19 \times 10^8 - 1.12 \times 10^8}{1.12 \times 10^8} \times 100$	2
Expresses as % difference: $6.25\%$	1

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