SCSA Physics Science Inquiry Skills

2 sample questions with marking guides and sample answers · Avg. score: 57.3%

Q20
2022
VCAA
1 mark
Q20
1 mark

The experimental uncertainty of a measurement is best understood as

A

an estimate of the validity of the data.

B

a mistake in the experimental method used.

C

a mistake in the recording of a measurement.

D

an estimate of the maximum likely difference between the measurement and the true value.

Reveal Answer
A

an estimate of the validity of the data.

Validity refers to how well an experiment measures what it intends to measure, whereas uncertainty quantifies the precision and potential error margin of a specific measurement.

B

a mistake in the experimental method used.

Experimental uncertainty is an inherent part of any measurement process due to equipment limitations, not a procedural mistake or blunder.

C

a mistake in the recording of a measurement.

A mistake in recording data is a human error, whereas uncertainty represents the natural limits of precision in the measuring instrument or process.

D

an estimate of the maximum likely difference between the measurement and the true value.

Correct Answer

Experimental uncertainty provides a quantitative estimate of the range within which the true value of the measurement is expected to lie, accounting for inherent measurement limitations.

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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