How many SCSA Physics questions cover Science Understanding?

AusGrader has 358 SCSA Physics questions on Science Understanding, all with instant AI grading and detailed marking feedback.

SCSA (WACE) Physics — Science Understanding Questions & Answers | AusGrader

Q11

2023

QCAA

Paper 1

1 mark

Q11

1 mark

Coulomb’s law describes the observation that

A

an electromotive force in a circuit may be induced through changes in the magnetic flux.

B

charged particles moving across magnetic field lines experience a force.

C

a change in the electromotive force is opposed.

D

like electric charges repel one another.

Q10

2022

QCAA

Paper 1

1 mark

Q10

1 mark

Electric field strength refers to the

A

intensity of an electric field at a particular location.

B

change in electrical potential energy between two defined points.

C

sum of electrically charged particles passing a point in a given time.

D

physical property of an object experiencing a force in an electromagnetic field.

Q10

2023

SCSA

4 marks

Q10

4 marks

Estimate the de Broglie wavelength for a standard men's basketball travelling at 10.0 m s $^{-1}$ .

Using de Broglie's equation,

\begin{align*} \lambda &= \frac{h}{mv} \end{align*}

Taking the mass of a standard men's basketball as approximately

\begin{align*} m &= 0.62\ \text{kg} \end{align*}

then

\begin{align*} \lambda &= \frac{6.63\times10^{-34}}{(0.62)(10.0)} \\ &= 1.07\times10^{-34}\ \text{m} \end{align*}

So the estimated de Broglie wavelength is approximately

\begin{align*} \lambda &\approx 1.1\times10^{-34}\ \text{m} \end{align*}

Marking Criteria

Descriptor	Marks
Estimates mass of basketball	1
Substitutes $mv$ for $p$ in equation (using 0.60 kg)	1
Calculates answer	1
2 significant figures	1

Q12

2025

SCSA

5 marks

Q12

Scientists have discovered a new particle they call a 'pentaquark'. It consists of two up quarks, two down quarks and an anti-strange quark. They determined its mass to be $1.54\text{ GeV}/c^2$ .

Q12a

2 marks

Determine the overall charge of a pentaquark.

Charges on the quarks are : $2 \times \frac{2}{3}$ , $2 \times \frac{-1}{3}$ , $\frac{1}{3}$

Overall charge is $+1$

Marking Criteria

Descriptor	Marks
Correctly identifies charges on all quarks: $2 \times \frac{2}{3}$ , $2 \times \frac{-1}{3}$ , $\frac{1}{3}$	1
Adds to correct total: $+1$	1

Q12b

3 marks

Calculate the mass of a pentaquark in kilograms.

$m = \frac{E}{c^2} = \frac{(1.54 \times 10^9)(1.60 \times 10^{-19})}{(3.00 \times 10^8)^2} = 2.74 \times 10^{-27} \text{ kg}$

Marking Criteria

Descriptor	Marks
Uses $E = mc^2$ rearranged for $m$ : $m = \frac{E}{c^2}$	1
Substitutes values and converts GeV to Joules: $m = \frac{(1.54 \times 10^9)(1.60 \times 10^{-19})}{(3.00 \times 10^8)^2}$	1
Calculates answer: $m = 2.74 \times 10^{-27} \text{ kg}$	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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