How many SCSA Physics questions cover Science Inquiry Skills?

AusGrader has 69 SCSA Physics questions on Science Inquiry Skills, all with instant AI grading and detailed marking feedback.

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

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