Band 6

Matrices, GEN1 2024 VCAA 32 MC

A large sporting event is held over a period of four consecutive days: Thursday, Friday, Saturday and Sunday.

People can watch the event at four different sites throughout the city: Botanical Gardens $(G)$, City Square $(C)$, Riverbank $(R)$ or Main Beach $(M)$.

Let $S_n$ be the state matrix that shows the number of people at each location $n$ days after Thursday. The expected number of people at each location can be determined by the matrix recurrence rule

$S_{n+1}=TS_n+A$

\begin{aligned}
& \quad \quad \quad \quad \quad \quad \quad \quad \quad \textit{this day} \\
& \quad \quad \quad \quad \quad \quad \quad \ G \quad \ \ C \quad \ \ R \quad \ \ M \\
& \text{where} \quad T=\begin{bmatrix}
0.4 & 0.2 & 0.4 & 0 \\
0.4 & 0.1 & 0.3 & 0.3 \\
0.1 & 0.4 & 0.1 & 0.2 \\
0.1 & 0.3 & 0.2 & 0.5
\end{bmatrix}\begin{array}{l}
G \\
C \\
R \\
M
\end{array} \text { next day } \quad \text{and}& A=\begin{bmatrix}
300 \\
200\\
100 \\
300
\end{bmatrix}\begin{array}{l}
G \\
C \\
R \\
M
\end{array}
\end{aligned}

\begin{aligned} \text{Given the state matrix}& \quad \quad S_3=\begin{bmatrix}
5620\\
6386\\
4892\\
6902
\end{bmatrix}\begin{array}{l}
G \\
C \\
R \\
M
\end{array}
\end{aligned}

the number of people watching the event at the Botanical Gardens $(G)$ from Thursday to Sunday has

decreased by 162
decreased by 212
increased by 124
increased by 696

Show Answers Only

$D$

Show Worked Solution

$S_{n+1}$	$=TS_n+A$
$TS_n$	$=S_{n+1}-A$
$S_n$	$=T^{-1}\times \left(S_{n+1}-A\right)$

$\text{Using CAS:}$

$T =\begin{bmatrix}
\ 0.4 & 0.2 & 0.4 & 0 \\
\ 0.4 & 0.1 & 0.3 & 0.3 \\
\ 0.1 & 0.4 & 0.1 & 0.2 \\
\ 0.1 & 0.3 & 0.2 & 0.5
\end{bmatrix}\rightarrow T^{-1}=
\begin{bmatrix}
\ -2.6 & 5.4 & 3.4 & -4.6 \\
\ 0.2 & -0.8 & 3.2 & -0.8 \\
\ 5 & -5 & -5 & 5 \\
\ -1.6 & 1.4 & -0.6 & 1.4
\end{bmatrix}$

$\text{Sunday} =S_3=\begin{bmatrix}
\ 5620 \\
\ 6386 \\
\ 4892 \\
\ 6902
\end{bmatrix}\quad\text{Saturday} =S_2=T^{-1}\times \left(S_{3}-A\right)=\begin{bmatrix}
\ 5496 \\
\ 6168 \\
\ 4720 \\
\ 6516
\end{bmatrix} $

$\text{Friday} =S_1=T^{-1}\times \left(S_{2}-A\right)=\begin{bmatrix}
\ 5832 \\
\ 6076 \\
\ 4120 \\
\ 5972
\end{bmatrix} $

$\text{Thursday} =S_0=T^{-1}\times \left(S_{1}-A\right)=\begin{bmatrix}
\ 4924 \\
\ 4732 \\
\ 6540 \\
\ 4904
\end{bmatrix} $

$\therefore\ \text{The number of people attending the Botanical Gardens }$

$\text{has increased by}\ 5620-4924=696\ \ \text{from Thursday to Sunday.}$

$\Rightarrow D$

Recursion and Finance, GEN1 2024 VCAA 24 MC

André invested $18 000 in an account for five years, with interest compounding monthly.

He adds an extra payment into the account each month immediately after the interest is calculated.

For the first two years, the balance of the account, in dollars, after $n$ months, $A_n$, can be modelled by the recurrence relation

$A_0=18000, \quad A_{n+1}=1.002 A_n+100$

After two years, André decides he would like the account to reach a balance of $30 000 at the end of the five years.

He must increase the value of the monthly extra payment to achieve this.

The minimum value of the new payment for the last three years is closest to

$189.55
$195.45
$202.35
$246.55

Show Answers Only

$A$

Show Worked Solution

$\text{Step 1: Using CAS}$

$\text{Value of investment after 2 years}\ =$21\,340.18$

$\text{Step 2: Using CAS}$

$\text{Minimum value of new payment}\ =$189.55$

$\Rightarrow A$

Mechanics, EXT2 M1 2024 HSC 16c

Two particles, $A$ and $B$, each have mass 1 kg and are in a medium that exerts a resistance to motion equal to $k v$, where $k>0$ and $v$ is the velocity of any particle. Both particles maintain vertical trajectories.

The acceleration due to gravity is $g$ ms$^{-2}$, where $g>0$.

The two particles are simultaneously projected towards each other with the same speed, $v_0$ ms$^{-1}$, where $0<v_0<\dfrac{g}{k}$.

The particle $A$ is initially $d$ metres directly above particle $B$, where $d<\dfrac{2 v_0}{k}$.

Find the time taken for the particles to meet. (4 marks)

--- 16 WORK AREA LINES (style=lined) ---

Show Answers Only

$t=-\dfrac{1}{k} \ln \left(\dfrac{2v_0-dk}{v_0}\right)$

Show Worked Solution

$\text{Particle A:}$

$\ddot{x}=\dfrac{dv_{\small{A}}}{dt}=-g-k v_{\small{A}}$

$\dfrac{dt}{dv_{\small{A}}}=\dfrac{1}{-g-kv_{\small{A}}}$

$t=-\displaystyle \int \dfrac{1}{g+k v_{\small{A}}} \, dv_{\small{A}}=-\dfrac{1}{k} \ln \left(g+k v_A\right)+c$

$\text{At} \ \ t=0, \ v_{\small{A}}=-v_0 \ \Rightarrow \ c=\dfrac{1}{k} \ln \left(g-k v_0\right)$

$t=\dfrac{1}{k} \ln \left(g-k v_0\right)-\dfrac{1}{k} \ln \left(g+kv_{\small{A}}\right)=\dfrac{1}{k} \ln \left(\dfrac{g-kv_0}{g+kv_{\small{A}}}\right)$

$\text{Find} \ v_{\small{A}}:$

	$\dfrac{g-k v_0}{g+k v_{\small{A}}}$	$=e^{kt}$
	$g-kv_0$	$=e^{kt} \cdot g+e^{kt} \cdot kv_{\small{A}}$
	$kv_{\small{A}}$	$=e^{-kt}\left(g-kv_0\right)-g$
	$v_{\small{A}}$	$=\dfrac{1}{k}\left[e^{-kt}\left(g-kv_0\right)-g\right]$

	$x$	$=\displaystyle \frac{1}{k} \int e^{-kt} \cdot g-e^{-kt} \cdot v_0-g \, dt$
		$=\dfrac{1}{k}\left[-\dfrac{g}{k}e^{-kt}+v_0 e^{-kt}-gt\right]+c$
		$=-\left(\dfrac{g}{k^2}-\dfrac{v_0}{k}\right) e^{-kt}-gt+c$

$\text{When} \ \ t=0, x=0 \ \Rightarrow \ c=\dfrac{g}{k^2}-\dfrac{v_0}{k}$

$x_{\small{A}}=d-\left(\dfrac{g}{k^2}-\dfrac{v_0}{k}\right) e^{-kt}-gt+\dfrac{g}{k^2}-\dfrac{v_0}{k}$

$\text{Particle B:}$

$\ddot{x}=-g-k v_{\small{B}}$

$t=-\dfrac{1}{k} \ln \left(g+k v_{\small{B}}\right)+c$

$\text{When} \ \ t=0, v_B=v_0 \ \Rightarrow \ c=\dfrac{1}{k} \ln \left(g+kv_0\right)$

	$v_{\small{B}}$	$=\dfrac{1}{k}\left[e^{-kt}\left(g+kv_0\right)-g\right]$
	$x_{\small{B}}$	$=-\left(\dfrac{g}{k^2}+\dfrac{v_0}{k}\right) e^{-kt}-gt+c$

$\text{When} \ \ t=0, x=0 \ \Rightarrow \ c=\dfrac{g}{k^2}+\dfrac{v_0}{k}$

$x_{\small{B}}=-\left(\dfrac{g}{k^2}+\dfrac{v_0}{k}\right) e^{-kt}-gt+\dfrac{g}{k^2}+\dfrac{v_0}{k}$

$\text{Find \(t$ when $\ x_{\small{A}}=x_{\small{B}}$:}\)

$d-\left(\dfrac{g}{k^2}-\dfrac{v_0}{k}\right) e^{-k t}-g t+\dfrac{g}{k^2}-\dfrac{v_0}{k}=-\left(\dfrac{g}{k^2}+\dfrac{v_0}{k}\right) e^{-k t}-g t+\dfrac{g}{k^2}+\dfrac{v_0}{k}$

	$\dfrac{2 v_0}{k} \cdot e^{-k t}$	$=\dfrac{2 v_0}{k}-d$
	$e^{-kt}$	$=\left(\dfrac{2 v_0-d k}{k}\right) \cdot \dfrac{k}{v_0}$
	$-kt$	$=\ln \left(\dfrac{2 v_0-d k}{v_0}\right)$
	$t$	$=-\dfrac{1}{k} \ln \left(\dfrac{2v_0-dk}{v_0}\right)$

Complex Numbers, EXT2 N2 2024 HSC 16b

The number $w=e^{\small{\dfrac{2 \pi i}{3}}}$ is a complex cube root of unity. The number $\gamma$ is a cube root of $w$.

Show that $\gamma+\bar{\gamma}$ is a real root of $z^3-3 z+1=0$. (3 marks)

--- 12 WORK AREA LINES (style=lined) ---
By using part (i) to find the exact value of $\cos \dfrac{2 \pi}{9} \cos \dfrac{4 \pi}{9} \cos \dfrac{8 \pi}{9}$, deduce the value(s) of $\cos \dfrac{2^n \pi}{9} \cos \dfrac{2^{n+1} \pi}{9} \cos \dfrac{2^{n+2} \pi}{9}$ for all integers $n \geq 1$. Justify your answer. (3 marks)

--- 12 WORK AREA LINES (style=lined) ---

Show Answers Only

i. $w=e^{\small{\dfrac{2 \pi i}{3}}} \Rightarrow \ \text{complex (cube) root of 1}$

$\gamma^3=w, \quad \gamma \bar{\gamma}=\abs{\gamma}^2=1$

$\text{Show}\ \ \gamma+\bar{\gamma}\ \ \text{is a real root of} \ \ z^3-3 z+1=0:$

	$(\gamma+\bar{\gamma})^3-3(\gamma+\bar{\gamma})+1$
		$=\gamma^3+3 \gamma^2 \bar{\gamma}+3 \gamma \bar{\gamma}^2+\bar{\gamma}^3-3 \gamma-3 \bar{\gamma}+1$
		$=\gamma^3+3 \gamma\abs{\gamma}^2+3 \bar{\gamma}\abs{\gamma}^2+\bar{\gamma}^3-3 \gamma-3 \bar{\gamma}+1$
		$=w+3 \gamma+3 \bar{\gamma}+\bar{w}-3 \gamma-3 \bar{\gamma}+1$
		$=w+\bar{w}+1$
		$=0 \quad \text{(Sum of complex roots of 1 = 0)}$

$\text{Show}\ \ \gamma+\bar{\gamma} \ \ \text{is real:}$

$\gamma+\bar{\gamma}=a+b i+a-b i=2 a \quad(a \in \mathbb{R})$

$\therefore \gamma+\bar{\gamma} \ \ \text{is a real root of} \ \ z^3-3 z+1=0.$

ii. $e^{\small{\dfrac{2 \pi i}{9}}}\ \ \text{is a cube root of}\ \ \omega \ \Rightarrow \ \Bigg(e^{\small{\dfrac{2 \pi i}{9}}}\Bigg)^3=e^{\small{\dfrac{2 \pi i}{3}}}$

$\text{Cubic roots are} \ \dfrac{2}{3} \ \text {rotations from each other.}$

$\text{Roots of} \ \ z^3-3 z+1=0:$

	$(\gamma+\bar{\gamma})_1$	$=2 \cos \left(\dfrac{2 \pi}{9}\right)$
	$(\gamma+\bar{\gamma})_2$	$=2 \cos \left(\dfrac{8 \pi}{9}\right)$
	$(\gamma+\bar{\gamma})_3$	$=2 \cos \left(\dfrac{4\pi}{9}\right)$

$\text{Product of roots} \ \ \alpha B \gamma=-\dfrac{d}{a}:$

	$2 \cos \left(\dfrac{2 \pi}{9}\right) \cdot 2 \cos \left(\dfrac{8 \pi}{9}\right) \cdot 2 \cos \left(\dfrac{4 \pi}{4}\right)$	$=-1$
	$\cos \left(\dfrac{2 \pi}{9}\right) \cdot \cos \left(\dfrac{8 \pi}{9}\right) \cdot \cos \left(\dfrac{4 \pi}{9}\right)$	$=-\dfrac{1}{8}$

$\text{Consider} \ \ \cos \left(\dfrac{2 \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+1} \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+2} \pi}{9}\right):$

$\text{If} \ \ n=1,$

$\cos \left(\dfrac{2 \pi}{9}\right) \cdot \cos \left(\dfrac{4 \pi}{9}\right) \cdot \cos \left(\dfrac{8 \pi}{9}\right)=-\dfrac{1}{8}\ \ \text{(see above)}$

$\text {If} \ \ n=2,$

$\cos \left(\dfrac{4 \pi}{9}\right) \cdot \cos \left(\dfrac{8 \pi}{9}\right) \cdot \cos \left(\dfrac{2\pi}{9}\right)=-\dfrac{1}{8}$

$\Rightarrow \ \text{Multiplying each argument by 2 does not change the equation}$

$\therefore \cos \left(\dfrac{2^n \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+1} \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+2} \pi}{9}\right)=-\dfrac{1}{8}$

Show Worked Solution

i. $w=e^{\small{\dfrac{2 \pi i}{3}}} \Rightarrow \ \text{complex (cube) root of 1}$

$\gamma^3=w, \quad \gamma \bar{\gamma}=\abs{\gamma}^2=1$

$\text{Show}\ \ \gamma+\bar{\gamma}\ \ \text{is a real root of} \ \ z^3-3 z+1=0:$

	$(\gamma+\bar{\gamma})^3-3(\gamma+\bar{\gamma})+1$
		$=\gamma^3+3 \gamma^2 \bar{\gamma}+3 \gamma \bar{\gamma}^2+\bar{\gamma}^3-3 \gamma-3 \bar{\gamma}+1$
		$=\gamma^3+3 \gamma\abs{\gamma}^2+3 \bar{\gamma}\abs{\gamma}^2+\bar{\gamma}^3-3 \gamma-3 \bar{\gamma}+1$
		$=w+3 \gamma+3 \bar{\gamma}+\bar{w}-3 \gamma-3 \bar{\gamma}+1$
		$=w+\bar{w}+1$
		$=0 \quad \text{(Sum of complex roots of 1 = 0)}$

$\text{Show}\ \ \gamma+\bar{\gamma} \ \ \text{is real:}$

$\gamma+\bar{\gamma}=a+b i+a-b i=2 a \quad(a \in \mathbb{R})$

$\therefore \gamma+\bar{\gamma} \ \ \text{is a real root of} \ \ z^3-3 z+1=0.$

ii. $e^{\small{\dfrac{2 \pi i}{9}}}\ \ \text{is a cube root of}\ \ \omega \ \Rightarrow \ \Bigg(e^{\small{\dfrac{2 \pi i}{9}}}\Bigg)^3=e^{\small{\dfrac{2 \pi i}{3}}}$

$\text{Cubic roots are} \ \dfrac{2}{3} \ \text {rotations from each other.}$

$\text{Roots of} \ \ z^3-3 z+1=0:$

	$(\gamma+\bar{\gamma})_1$	$=2 \cos \left(\dfrac{2 \pi}{9}\right)$
	$(\gamma+\bar{\gamma})_2$	$=2 \cos \left(\dfrac{8 \pi}{9}\right)$
	$(\gamma+\bar{\gamma})_3$	$=2 \cos \left(\dfrac{4\pi}{9}\right)$

$\text{Product of roots} \ \ \alpha B \gamma=-\dfrac{d}{a}:$

	$2 \cos \left(\dfrac{2 \pi}{9}\right) \cdot 2 \cos \left(\dfrac{8 \pi}{9}\right) \cdot 2 \cos \left(\dfrac{4 \pi}{4}\right)$	$=-1$
	$\cos \left(\dfrac{2 \pi}{9}\right) \cdot \cos \left(\dfrac{8 \pi}{9}\right) \cdot \cos \left(\dfrac{4 \pi}{9}\right)$	$=-\dfrac{1}{8}$

$\text{Consider} \ \ \cos \left(\dfrac{2 \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+1} \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+2} \pi}{9}\right):$

$\text{If} \ \ n=1,$

$\cos \left(\dfrac{2 \pi}{9}\right) \cdot \cos \left(\dfrac{4 \pi}{9}\right) \cdot \cos \left(\dfrac{8 \pi}{9}\right)=-\dfrac{1}{8}\ \ \text{(see above)}$

$\text {If} \ \ n=2,$

$\cos \left(\dfrac{4 \pi}{9}\right) \cdot \cos \left(\dfrac{8 \pi}{9}\right) \cdot \cos \left(\dfrac{2\pi}{9}\right)=-\dfrac{1}{8}$

$\Rightarrow \ \text{Multiplying each argument by 2 does not change the equation}$

$\therefore \cos \left(\dfrac{2^n \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+1} \pi}{9}\right) \cdot \cos \left(\dfrac{2^{n+2} \pi}{9}\right)=-\dfrac{1}{8}$

Calculus, EXT2 C1 2024 HSC 15d

Using a suitable substitution, find $\displaystyle\int \dfrac{2 x^2}{\sqrt{2 x-x^2}}\, d x$. (3 marks)

--- 5 WORK AREA LINES (style=lined) ---

Show Answers Only

$I=3 \sin ^{-1}(x-1)-4 \sqrt{2x-x^2}-(x-1) \sqrt{2x-x^2}+c$

Show Worked Solution

	$\displaystyle \int \dfrac{2x^2}{\sqrt{2x-x^2}}\, dx$	$=\displaystyle \int \frac{2 x^2}{\sqrt{1-(1-2 x+x^2)}}\, dx$
		$=\displaystyle \int \frac{2 x^2}{\sqrt{1-(x-1)^2}} \, dx$

$\text {Let} \ \ x-1=\sin \theta \ \Rightarrow \ x=1+\sin \theta$

$\dfrac{dx}{d \theta}=\cos \theta \ \Rightarrow \ dx=\cos \theta \, d \theta$

	$I$	$=\displaystyle \int \frac{2(1+\sin \theta)^2}{\sqrt{1-\sin ^2 \theta}} \cdot \cos \theta \, d \theta$
		$=2 \displaystyle \int \frac{1+2 \sin \theta+\sin ^2 \theta}{\cos \theta} \cdot \cos \theta \, d \theta$
		$=2 \displaystyle \int 1+2 \sin \theta+\frac{1}{2}(1-\cos (2 \theta)) \, d \theta$
		$=\displaystyle \int 2+4 \sin \theta+1-\cos (2 \theta) \, d \theta$
		$=\displaystyle \int 3+4 \sin \theta-\cos (2 \theta) \, d \theta$
		$=3 \theta-4 \cos \theta-\dfrac{1}{2} \sin (2 \theta)+c$
		$=3 \theta-4 \cos \theta-\sin \theta \cos \theta+c$

$\Rightarrow \cos \theta=\sqrt{1^2-(x-1)^2}=\sqrt{2x-x^2}$

$\therefore I=3 \sin ^{-1}(x-1)-4 \sqrt{2x-x^2}-(x-1) \sqrt{2x-x^2}+c$

Vectors, EXT2 V1 2024 HSC 14e

The diagram shows triangle $O Q A$.

The point $P$ lies on $O A$ so that $O P: O A=3: 5$.

The point $B$ lies on $O Q$ so that $O B: O Q=1: 3$.

The point $R$ is the intersection of $A B$ and $P Q$.

The point $T$ is chosen on $A Q$ so that $O, R$ and $T$ are collinear.

Let $\underset{\sim}{a}=\overrightarrow{O A}, \ \underset{\sim}{b}=\overrightarrow{O B}$ and $\overrightarrow{P R}=k \overrightarrow{P Q}$ where $k$ is a real number.

Show that $\overrightarrow{O R}=\dfrac{3}{5}(1-k) \underset{\sim}{a}+3 k \underset{\sim}{b}$. (2 marks)

--- 5 WORK AREA LINES (style=lined) ---

Writing $\overrightarrow{A R}=h \overrightarrow{A B}$, where $h$ is a real number, it can be shown that $\overrightarrow{O R}=(1-h) \underset{\sim}{a}+h \underset{\sim}{b}$. (Do NOT prove this.)

Show that $k=\dfrac{1}{6}$. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---
Find $\overrightarrow{O T}$ in terms of $\underset{\sim}{a}$ and $\underset{\sim}{b}$. (2 marks)

--- 4 WORK AREA LINES (style=lined) ---

Show Answers Only

i. $\underset{\sim}{a}=\overrightarrow{O A}, \ \underset{\sim}{b}=\overrightarrow{O B}, \ \overrightarrow{P R}=k \overrightarrow{P Q}$

$\overrightarrow{O P}: \overrightarrow{O A}=3: 5 \ \Rightarrow \ \overrightarrow{O P} = \dfrac{3}{5} \times \overrightarrow{O A} $

$\overrightarrow{O B}: \overrightarrow{O Q}=1: 3\ \Rightarrow \ \overrightarrow{O Q} = 3 \times \overrightarrow{OB} $

$\text {Show } \overrightarrow{O R}=\dfrac{3}{5}(1-k) \underset{\sim}{a}+3 k \underset{\sim}{b}$

	$\overrightarrow{O R}$	$=\overrightarrow{O P}+k \overrightarrow{P Q}$
		$=\overrightarrow{O P}+k(\overrightarrow{O Q}-\overrightarrow{O P})$
		$=(1-k) \overrightarrow{O P}+k \overrightarrow{O Q}$
		$=\dfrac{3}{5}(1-k)\underset{\sim}{a}+3 k \underset{\sim}{b}$

ii. $\overrightarrow{A R}=h \overrightarrow{A B}, \quad h \in \mathbb{R}$

$\overrightarrow{OR}=(1-h) \underset{\sim}{a}+h \underset{\sim}{b} \ \ \text{(given)}$

$\overrightarrow{O R}=\dfrac{3}{5}(1-k) \underset{\sim}{a}+3 k \underset{\sim}{b} \ \ \text{(part(i))}$

$\text{Vector }\underset{\sim}{a} \neq \lambda \underset{\sim}{b}\ (\lambda \in \mathbb{R})\ \Rightarrow \ \text{linearly independent and are basis vectors for } \overrightarrow{O R}$.

$\text {Equating coefficients:}$

$\dfrac{3}{5}(1-k)=1-h\ \ldots\ (1)$

$h=3 k\ \ldots\ (2)$

$\text{Substituting (2) into (1)}$

	$\dfrac{3}{5}(1-k)$	$=1-3 k$
	$3-3 k$	$=5-15 k$
	$k$	$=\dfrac{1}{6}$

iii. $\overrightarrow{O T}=\dfrac{3}{4}(\underset{\sim}{a}+\underset{\sim}{b})$

Show Worked Solution

i. $\underset{\sim}{a}=\overrightarrow{O A}, \ \underset{\sim}{b}=\overrightarrow{O B}, \ \overrightarrow{P R}=k \overrightarrow{P Q}$

$\overrightarrow{O P}: \overrightarrow{O A}=3: 5 \ \Rightarrow \ \overrightarrow{O P} = \dfrac{3}{5} \times \overrightarrow{O A} $

$\overrightarrow{O B}: \overrightarrow{O Q}=1: 3\ \Rightarrow \ \overrightarrow{O Q} = 3 \times \overrightarrow{OB} $

$\text {Show } \overrightarrow{O R}=\dfrac{3}{5}(1-k) \underset{\sim}{a}+3 k \underset{\sim}{b}$

	$\overrightarrow{O R}$	$=\overrightarrow{O P}+k \overrightarrow{P Q}$
		$=\overrightarrow{O P}+k(\overrightarrow{O Q}-\overrightarrow{O P})$
		$=(1-k) \overrightarrow{O P}+k \overrightarrow{O Q}$
		$=\dfrac{3}{5}(1-k)\underset{\sim}{a}+3 k \underset{\sim}{b}$

ii. $\overrightarrow{A R}=h \overrightarrow{A B}, \quad h \in \mathbb{R}$

$\overrightarrow{OR}=(1-h) \underset{\sim}{a}+h \underset{\sim}{b} \ \ \text{(given)}$

$\overrightarrow{O R}=\dfrac{3}{5}(1-k) \underset{\sim}{a}+3 k \underset{\sim}{b} \ \ \text{(part(i))}$

$\text{Vector }\underset{\sim}{a} \neq \lambda \underset{\sim}{b}\ (\lambda \in \mathbb{R})\ \Rightarrow \ \text{linearly independent and are basis vectors for } \overrightarrow{O R}$.

$\text {Equating coefficients:}$

$\dfrac{3}{5}(1-k)=1-h\ \ldots\ (1)$

$h=3 k\ \ldots\ (2)$

$\text{Substituting (2) into (1)}$

	$\dfrac{3}{5}(1-k)$	$=1-3 k$
	$3-3 k$	$=5-15 k$
	$k$	$=\dfrac{1}{6}$

iii. $\overrightarrow{O T}=\lambda \overrightarrow{O R}$

$\text {Using parts (i) and (ii):}$

$\overrightarrow{O R}=\dfrac{3}{5}\left(1-\dfrac{1}{6}\right)\underset{\sim}{a}+3\left(\dfrac{1}{6}\right) \underset{\sim}{b}=\dfrac{1}{2}\left(\underset{\sim}{a}+\underset{\sim}{b}\right)$

$\overrightarrow{O T}=\dfrac{\lambda}{2}(\underset{\sim}{a}+\underset{\sim}{b})$

$\text {Find } \lambda:$

	$\overrightarrow{O T}$	$=\overrightarrow{O A}+\mu \overrightarrow{A Q}$
	$\dfrac{\lambda}{2}(\underset{\sim}{a}+\underset{\sim}{b})$	$=\underset{\sim}{a}+\mu(3 \underset{\sim}{b}-\underset{\sim}{a}) \quad \Big(\text{noting}\ \overrightarrow{A Q}=\overrightarrow{O Q}-\overrightarrow{O A}=3 \underset{\sim}{b}-\underset{\sim}{a}\Big)$
		$=\underset{\sim}{a}(1-\mu)+3 \mu \underset{\sim}{b}$

$\text {Equating coefficients:}$

$\dfrac{\lambda}{2}=3 \mu \ \Rightarrow \ \mu=\dfrac{\lambda}{6}$

	$1-\dfrac{\lambda}{6}$	$=\dfrac{\lambda}{2}$
	$6-\lambda$	$=3 \lambda$
	$\lambda$	$=\dfrac{3}{2}$

$\therefore \overrightarrow{O T}=\dfrac{3}{4}(\underset{\sim}{a}+\underset{\sim}{b})$

Proof, EXT2 P1 2024 HSC 13d

It is known that for all positive real numbers $x, y$

$x+y \geq 2 \sqrt{x y} .$ (Do NOT prove this.)

Show that if $a, b, c$ are positive real numbers with $\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1$ then $a \sqrt{b c}+b \sqrt{a c}+c \sqrt{a b} \leq a b c$. (3 marks)

--- 10 WORK AREA LINES (style=lined) ---

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$\text{If }\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1, \ \ \ a, b, c \in \mathbb{R} ^{+}$

$\text {Prove} \ \ a \sqrt{b c}+b \sqrt{a c}+c \sqrt{a b} \leqslant a b c$

$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$	$=1$
$\dfrac{b c+a c+a b}{a b c}$	$=1$
$ab+bc+ac$	$=abc\ …\ (1)$

$x+y \geqslant 2 \sqrt{x y}, \quad x, y \in \mathbb{R}^{+} \text{(given)}$

$a+b \geqslant 2 \sqrt{a b} \ \Rightarrow \ \sqrt{a b} \leqslant \dfrac{1}{2}(a+b)$

$\text{Similarly,}$

$\sqrt{bc} \leqslant \dfrac{1}{2}(b+c) \ \text{and}\ \ \sqrt{a c} \leqslant \dfrac{1}{2}(a+c)$

	$a \sqrt{b c}+b \sqrt{a c}+c \sqrt{a b}$	$\leqslant a \times \dfrac{1}{2}(b+c)+b \times \dfrac{1}{2}(a+c)+c \times \dfrac{1}{2}(a+b)$
		$\leqslant \dfrac{1}{2} a b+\dfrac{1}{2} a c+\dfrac{1}{2} a b+\dfrac{1}{2} b c+\dfrac{1}{2} a c+\dfrac{1}{2} b c$
		$\leqslant a b+b c+a c$
		$\leqslant a b c\ \ \text{(see (1) above)}$

Show Worked Solution

$\text{If }\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}=1, \ \ \ a, b, c \in \mathbb{R} ^{+}$

$\text {Prove} \ \ a \sqrt{b c}+b \sqrt{a c}+c \sqrt{a b} \leqslant a b c$

$\dfrac{1}{a}+\dfrac{1}{b}+\dfrac{1}{c}$	$=1$
$\dfrac{b c+a c+a b}{a b c}$	$=1$
$ab+bc+ac$	$=abc\ …\ (1)$

$x+y \geqslant 2 \sqrt{x y}, \quad x, y \in \mathbb{R}^{+} \text{(given)}$

$a+b \geqslant 2 \sqrt{a b} \ \Rightarrow \ \sqrt{a b} \leqslant \dfrac{1}{2}(a+b)$

$\text{Similarly,}$

$\sqrt{bc} \leqslant \dfrac{1}{2}(b+c) \ \text{and}\ \ \sqrt{a c} \leqslant \dfrac{1}{2}(a+c)$

	$a \sqrt{b c}+b \sqrt{a c}+c \sqrt{a b}$	$\leqslant a \times \dfrac{1}{2}(b+c)+b \times \dfrac{1}{2}(a+c)+c \times \dfrac{1}{2}(a+b)$
		$\leqslant \dfrac{1}{2} a b+\dfrac{1}{2} a c+\dfrac{1}{2} a b+\dfrac{1}{2} b c+\dfrac{1}{2} a c+\dfrac{1}{2} b c$
		$\leqslant a b+b c+a c$
		$\leqslant a b c\ \ \text{(see (1) above)}$

Vectors, EXT1 V1 2024 HSC 14d

A particle is projected from the origin, with initial speed $V$ at an angle of $\theta$ to the horizontal. The position vector of the particle, $\underset{\sim}{r}(t)$, where $t$ is the time after projection and $g$ is the acceleration due to gravity, is given by

$\underset{\sim}{r}(t)=\left(\begin{array}{c}Vt\cos\theta \\Vt\sin \theta -\dfrac{gt^2}{2}\end{array}\right)$. (Do NOT prove this.)

Let $D(t)$ be the distance of the particle from the origin at time $t$, so $D(t)=|\underset{\sim}{r}(t)|$.

Show that for $\theta<\sin ^{-1}\left(\sqrt{\dfrac{8}{9}}\right)$ the distance, $D(t)$, is increasing for all $t>0$. (4 marks)

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Show Answers Only

$\text{See Worked Solutions.}$

Show Worked Solution

$\underset{\sim}{r}(t)=\left(\begin{array}{c}Vt\cos\theta \\Vt\sin \theta -\dfrac{gt^2}{2}\end{array}\right), \ D(t)=\abs{\underset{\sim}{r}(t)}$

	$D(t)^2$	$=(V t \cos \theta)^2+\left(V t \sin \theta-\dfrac{g t^2}{2}\right)^2$
		$=V^2 t^2 \cos ^2 \theta+V^2 t^2 \sin ^2 \theta-V g t^3 \sin \theta+\dfrac{g^2 t^4}{4}$
		$=V^2 t^2\left(\cos ^2 \theta+\sin ^2 \theta\right)-V g t^3 \sin \theta+\dfrac{g^2 t^4}{4}$
		$=V^2 t^2-V g t^3 \sin \theta+\dfrac{g^2 t^4}{4}$

$\text{Let }F(t)=D(t)^2$

	$F^{\prime}(t)$	$=2 V^2 t-3 V g t^2 \sin \theta+g^2 t^3$
		$=t\left(2 V^2-3 V g t \sin \theta+g^2 t^2\right)$

$\text {Monotonically increasing when } F^{\prime}(t)>0 \text { for all } t.$

$g^2 t^2-3 V g t\, \sin \theta+2 V^2>0$

$\text {Find } t \text { when } \Delta<0:$

	$(-3 V g \sin \theta)^2-4 \cdot g^2 \cdot 2 V^2$	$<0$
	$9 V^2 g^2 \sin ^2 \theta-8 V^2 g^2$	$<0$
	$\sin ^2 \theta$	$<\dfrac{8}{9}$
	$\sin \theta$	$<\sqrt{\dfrac{8}{9}} \ \ \left(\theta \in\left(0, \dfrac{\pi}{2}\right) \Rightarrow \ \sin \theta \in(0,1)\right)$
	$\theta$	$<\sin ^{-1}\left(\sqrt{\dfrac{8}{9}}\right)$

Trigonometry, EXT1 T3 2024 HSC 10 MC

For real numbers $a$ and $b$, where $a \neq 0$ and $b \neq 0$, we can find numbers $\alpha$, $\beta$, $\gamma$, $\delta$ and $R$ such that $a\,\cos x + b\,\sin x$ can be written in the following 4 forms:

$R\,\sin(x + \alpha)$

$R\,\sin(x-\beta)$

$R\,\cos(x + \gamma)$

$R\,\cos(x-\delta)$

where $R \gt 0$ and $0<\alpha, \beta, \gamma, \delta \lt 2\pi$.

What is the value of $\alpha + \beta + \gamma + \delta$?

$0$
$\pi$
$2\pi$
$4\pi$

Show Answers Only

$D$

Show Worked Solution

$a\,\sin\,x+b\,\cos\,x\ \ \text{can be written 4 ways.}$

$\text{Consider the case:}$

$\cos\,x+\sin\,x\ \ \text{where}\ a=b=1,\ \ R=\sqrt{2}$

$\sqrt{2}\,\sin(x+\alpha)$	$= \sqrt{2}(\sin\,x\,\cos\,\alpha + \cos\,x\,\sin\,\alpha)$
	$= \sqrt{2}\Big(\sin\,x \cdot \dfrac{1}{\sqrt2} + \cos\,x \cdot \dfrac{1}{\sqrt2}\Big)$

$\cos\, \alpha^{+}, \sin\,\alpha^{+}\ \Rightarrow\ \ \alpha=\dfrac{\pi}{4} $

$\text{Similarly for other 3 cases:}$

$\sqrt{2}\,\sin(x-\beta): \ \cos\, \beta^{+}, \sin\,\beta^{-}\ \Rightarrow\ \ \beta=\dfrac{7\pi}{4} $

$\sqrt{2}\,\cos(x+\gamma): \ \cos\, \gamma^{+}, \sin\,\gamma^{-}\ \Rightarrow\ \ \gamma=\dfrac{7\pi}{4} $

$\sqrt{2}\,\cos(x-\delta): \ \cos\, \delta^{+}, \sin\,\delta^{+}\ \Rightarrow\ \ \delta=\dfrac{\pi}{4} $

$\therefore \alpha + \beta + \gamma + \delta = 4\pi$

$\Rightarrow D$

Trigonometry, EXT1 T1 2024 HSC 14c

Explain why the equation $\tan ^{-1}(3 x)+\tan ^{-1}(10 x)=\theta$, where $-\pi<\theta<\pi$, has exactly one solution. (1 marks)

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Solve $\tan ^{-1}(3 x)+\tan ^{-1}(10 x)=\dfrac{3 \pi}{4}$. (2 marks)

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i. $\tan ^{-1}(3 x)+\tan ^{-1}(10 x)=\theta \quad-\pi<\theta<\pi$

$\text {Range:}\ \ \tan ^{-1}(3 x) \in\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right), \ \tan ^{-1}(10 x) \in\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$

$\Rightarrow \text { Both are monotonically increasing functions}$

$\Rightarrow\tan ^{-1}(3 x)+\tan ^{-1}(10 x) \text{ is also monotonically increasing with range }(-\pi, \pi)$

$\Rightarrow \text{ Only 1 solution exists (horizontal line will only cut graph once).}$

ii. $x=\dfrac{1}{2}$

Show Worked Solution

i. $\tan ^{-1}(3 x)+\tan ^{-1}(10 x)=\theta \quad-\pi<\theta<\pi$

$\text {Range:}\ \ \tan ^{-1}(3 x) \in\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right), \ \tan ^{-1}(10 x) \in\left(-\dfrac{\pi}{2}, \dfrac{\pi}{2}\right)$

$\Rightarrow \text { Both are monotonically increasing functions}$

$\Rightarrow\tan ^{-1}(3 x)+\tan ^{-1}(10 x) \text{ is also monotonically increasing with range }(-\pi, \pi)$

$\Rightarrow \text{ Only 1 solution exists (horizontal line will only cut graph once).}$

ii. $\tan ^{-1}(3 x)+\tan ^{-1}(10 x)=\dfrac{3 \pi}{4}$

$\tan \left(\tan ^{-1}(3 x)+\tan ^{-1}(10 x)\right)=\tan \left(\dfrac{3 \pi}{4}\right)$

$\dfrac{\tan \left(\tan ^{-1}(3 x)\right)+\tan \left(\tan ^{-1}(10 x)\right)}{1-\tan \left(\tan ^{-1}(3 x)\right) \cdot \tan \left(\tan ^{-1}(10 x)\right)}=-1$

	$\dfrac{3 x+10 x}{1-30 x^2}$	$=-1$
	$13 x$	$=30 x^2-1$
	$30 x^2-13 x-1$	$=0$
	$(15 x+1)(2 x-1)$	$=0$

$x=\dfrac{1}{2}\ \ \text {or}\ \ -\dfrac{1}{15}$

$\text {Graph is monotonically increasing through } (0,0) \Rightarrow \ \Big(x \neq -\dfrac{1}{15} \Big)$

$\therefore x=\dfrac{1}{2}$

Vectors, EXT1 V1 2024 HSC 13c

The vector $\underset{\sim}{a}$ is $\displaystyle \binom{1}{3}$ and the vector $\underset{\sim}{b}$ is $\displaystyle\binom{2}{-1}$.

The projection of a vector $\underset{\sim}{x}$ onto the vector $\underset{\sim}{a}$ is $k \underset{\sim}{a}$, where $k$ is a real number.

The projection of the vector $\underset{\sim}{x}$ onto the vector $\underset{\sim}{b}$ is $p \underset{\sim}{b}$, where $p$ is a real number.

Find the vector $\underset{\sim}{x}$ in terms of $k$ and $p$. (4 marks)

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$\underset{\sim}{x}=\dfrac{5}{7} \displaystyle \binom{2 k+3 p}{4 k-p}$

Show Worked Solution

$\underset{\sim}{a}=\displaystyle \binom{1}{3}, \ \abs{\underset{\sim}{a}}=\sqrt{1^2+3^2}=\sqrt{10}$

$\underset{\sim}{b}=\displaystyle \binom{2}{-1}, \ \abs{\underset{\sim}{b}}=\sqrt{2^2+(-1)^2}=\sqrt{5}$

$\text{Let } \underset{\sim}{x}=\displaystyle \binom{x_1}{x_2}$

$\operatorname{proj}_{\underset{\sim}{a}} \underset{\sim}{x}=\dfrac{\underset{\sim}{x} \cdot \underset{\sim}{a}}{|\underset{\sim}{a}|^2} \underset{\sim}{a}=\dfrac{x_1+3 x_2}{10} \cdot \underset{\sim}{a}$

$k=\dfrac{x_1+3 x_2}{10} \ \Rightarrow \ x_1+3 x_2=10 k\ \ldots\ (1)$

$\operatorname{proj}_{\underset{\sim}{b}} \underset{\sim}{x}=\dfrac{\underset{\sim}{b} \cdot \underset{\sim}{x}}{|\underset{\sim}{b}|^2} b=\dfrac{2 x_1-x_2}{5} \cdot \underset{\sim}{b}$

$p=\dfrac{2 x_1-x_2}{5} \ \Rightarrow \ 2 x_1-x_2=5 p\ \ldots\\ (2)$

$\text {Multiply } (2) \times 3$

$6 x_1-3 x_2=15 p\ \ldots\ (3)$

$(1)+(3)$

$7 x_1$	$=10 k+15 p$
$x_1$	$=\dfrac{1}{7}(10 k+15)$

$\text {Multiply } (1) \times 2$

$2 x_1+6 x_2=20 k\ \ldots\ (4)$

$\text {Subtract} (4)-(2)$

$7x_2$	$=20 k-5 p$
$x_2$	$=\dfrac{1}{7}(20 k-5 p)$

$\therefore \underset{\sim}{x}=\displaystyle \frac{1}{7}\binom{10 k+15 p}{20 k-5 p}=\frac{5}{7}\binom{2 k+3 p}{4 k-p}$

CHEMISTRY, M2 EQ-Bank 10

A student is asked to prepare 500.0 mL of a 0.150 mol L$^{-1}$ standard solution of oxalic acid $\ce{(C2H2O4.2H2O)}$, and then to perform a dilution to produce 250.0 mL of a 0.0300 mol L$^{-1}$ solution. Outline and explain each step in this process, including the calculations involved and choice of equipment. (5 marks)

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Justify the procedure in part a by explaining two measures taken to ensure the accuracy of the standard solution and diluted solution produced. (2 marks)

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a. Calculate the mass of oxalic acid:

$MM\ce{(C2H2O4.2H2O)}= 2(12.01) + 2(1.008) + 4(16.00) + 4(1.008) +2(16.00) = 126.068\ \text{g mol}^{-1}$

$n\ce{(C2H2O4.2H2O)} = c \times V = 0.150 \times 0.5 = 0.075\ \text{mol}$

$m\ce{(C2H2O4.2H2O)} = 0.075 \times 126.068 = 9.455\ \text{g}$

→ 9.455 g of oxalic acid needs to be weighed using an electronic balance.

Prepare the standard solution:

→ Transfer the oxalic acid to a 500.0 mL volumetric flask using a funnel and rinse any remaining crystals from the weighing container into the flask with a small amount of distilled water.

→ Fill the volumetric flask with distilled water to about 80% full and swirl to dissolve the oxalic acid completely and carefully add more distilled water with a pipette until the bottom of the meniscus rests on the 500.0 mL mark to ensure precise volume.

→ Stopper the flask and invert several times to ensure a homogeneous solution.

Perform the dilution:

$V_1=\dfrac{c_2V_2}{c_1} = \dfrac{0.03 \times 250}{0.15} = 50\ \text{mL}$

→ Use a pipette to transfer 50.0 mL of the 0.150 mol L$^{-1}$ solution into a 250.0 mL volumetric flask.

→ Dilute with distilled water up to the 250.0 mL mark in the flask to achieve a concentration of 0.0300 mol L$^{-1}$.

b. Measures taken to ensure an accurate and precise solution:

→ A volumetric flask is used for both the standard solution and the diluted solution, as it provides precise measurements for the final solution volume. This accuracy is essential for ensuring the concentration is exactly as calculated.

→ The primary solute is weighed on an electronic balance to the nearest 0.01 g or better, minimizing any error in the amount of solute added to the solution.

→ The pipette provides precise measurements crucial for accurate dilutions.

→ Using oxalic acid as the primary standard for the investigation. Primary standard’s have high molar masses and are anhydrous (don’t absorb water). This ensures the substance is pure and the electronic balance can accurately weigh the sample.

Show Worked Solution

a. Calculate the mass of oxalic acid:

$MM\ce{(C2H2O4.2H2O)}= 2(12.01) + 2(1.008) + 4(16.00) + 4(1.008) +2(16.00) = 126.068\ \text{g mol}^{-1}$

$n\ce{(C2H2O4.2H2O)} = c \times V = 0.150 \times 0.5 = 0.075\ \text{mol}$

$m\ce{(C2H2O4.2H2O)} = 0.075 \times 126.068 = 9.455\ \text{g}$

→ 9.455 g of oxalic acid needs to be weighed using an electronic balance.

Prepare the standard solution:

→ Transfer the oxalic acid to a 500.0 mL volumetric flask using a funnel and rinse any remaining crystals from the weighing container into the flask with a small amount of distilled water.

→ Stopper the flask and invert several times to ensure a homogeneous solution.

Perform the dilution:

$V_1=\dfrac{c_2V_2}{c_1} = \dfrac{0.03 \times 250}{0.15} = 50\ \text{mL}$

→ Use a pipette to transfer 50.0 mL of the 0.150 mol L$^{-1}$ solution into a 250.0 mL volumetric flask.

→ Dilute with distilled water up to the 250.0 mL mark in the flask to achieve a concentration of 0.0300 mol L$^{-1}$.

b. Measures taken to ensure an accurate and precise solution:

→ The primary solute is weighed on an electronic balance to the nearest 0.01 g or better, minimizing any error in the amount of solute added to the solution.

→ The pipette provides precise measurements crucial for accurate dilutions.

Measurement, STD1 M5 2024 HSC 32

A scale diagram is shown with locations $A, B$ and $C$ marked (assume grid squares are 1 cm × 1 cm).

Jo takes 24 minutes to walk from $A$ to $B$ (in a straight line) when walking at 3 km per hour.

What is the scale used in the diagram? (3 marks)
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What is the distance from $B$ to $C$, in kilometres? (2 marks)
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a. $1:20\,000$

b. $\text{1.4 km}$

Show Worked Solution

a. $\text{Find distance }A\rightarrow B:$

$D=S \times T = 3 \times \dfrac{24}{60} = 1.2\ \text{km}$

	$\text{Scale }\rightarrow\ 6\ \text{cm }$	$:\ 1.2\ \text{km}$
	$6\ \text{cm }$	$:\ 1200\ \text{m}$
	$6\ \text{cm }$	$:\ 1200\times 100\ \text{cm}$
	$6\ $	$:\ 120\,000$
	$1\ $	$:\ 20\,000$

b. $\text{Distance }B\rightarrow C = 7\ \text{grid units}$

$\text{Real distance }B\rightarrow C$	$=7 \times 20\,000$
	$=140\,000\ \text{cm}$
	$=1400\ \text{m}$
	$=1.4\ \text{km}$

Algebra, STD1 A3 2024 HSC 23

Carrie is organising a fundraiser.

The cost of hiring the venue and the band is $2500. The cost of providing meals is $50 per person.

Complete the table of values to show the total cost of the fundraiser. (1 mark)

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\begin{array} {|l|c|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} \text{Number of people} \rule[-1ex]{0pt}{0pt} & \ \ \ \ 0\ \ \ \ & 25 & 50 & 75 & 100 & 125 & 150 \\
\hline
\rule{0pt}{2.5ex} \text{Cost} \rule[-1ex]{0pt}{0pt} & & 3750 & 5000 & 6250 & 7500 & 8750 & 10\,000 \\
\hline
\end{array}

Carrie decides that tickets should be sold at $70 per person. The graph shows the expected revenue at this ticket price. Using the information in part (a), plot the line that shows the cost of the fundraiser. (2 marks)
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How many tickets need to be sold for the fundraiser to break even? (1 mark)

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Carrie sold 300 tickets. How much profit did the fundraiser make? (3 marks)

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a. $\text{Cost when 0 people }= $2500$

c. $100\ \text{tickets}$

d. $\text{Profit }=$3500$

Show Worked Solution

a. $\text{Cost when 0 people }= $2500$

c. $\text{Point of intersection}\ \ \Rightarrow\ \ \text{break-even}$

$\therefore\ \text{Break-even when 100 tickets sold.}$

d. $\text{Revenue}\ (R)=70n\ \ (n=\ \text{number of people)}$

$\text{Cost}\ (C)=2500 + \Big(\dfrac{1250}{25}\Big)n=2500+50n$

$\text{Find profit}\ (P)\ \text{when}\ \ n=300:$

$P$	$=R-C$
	$=70\times 300-(2500+50\times300)$
	$=21\,000-17\,500$
	$=$3500$

Financial Maths, STD1 F3 2024 HSC 26

Bobby has a credit card that has no interest-free period.

Interest is charged at 0.07% per day, compounding daily, on the outstanding balance.

How much interest is Bobby charged on an outstanding balance of $600 for 30 days? (3 marks)

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$$12.73\ \text{(nearest cent)}$

Show Worked Solution

$PV=600,\ \ \ r=\dfrac{0.07}{100},\ \ \ n=30$

$FV$	$=PV(1+r)^n$
	$=600\Big(1+\dfrac{0.07}{100}\Big)^{30}$
	$=612.728\dots$

$\therefore\ \text{Interest}\ =$612.73-$600=$12.73\ \text{(nearest cent)}$

Measurement, STD1 M3 2024 HSC 14

A hotel is located 186 m north and 50 m west of a train station.

What is the straight line distance from the hotel to the train station? Round your answer to the nearest metre. (2 marks)

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What is the bearing of the hotel from the train station? Round your answer to the nearest degree. (2 marks)

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a. $193\ \text{m}$

b. $345^\circ\ \text{(nearest degree)}$

Show Worked Solution

a. $\text{By Pythagoras:}$

	$d^2$	$=50^2+186^2$
	$d^2$	$=37\,096$
	$d$	$=\sqrt{37\,096}$
		$=192.603\dots$
		$=193\ \text{m (nearest metre)}$

b.	$\tan\theta$	$=\dfrac{50}{186}$
	$\theta$	$=15.046\dots^\circ$
		$\approx 15^\circ\ \text{(nearest degree)}$

$\text{Bearing}\ H\ \text{from}\ T =360-15=345^\circ$

Algebra, STD1 A3 2024 HSC 10 MC

A cyclist rides a bicycle at a constant speed around a circular track.

Which of the graphs best illustrates the distance of the cyclist from the centre of the track as time varies?

Show Answers Only

$A$

Show Worked Solution

$\text{The bike rider is always a fixed distance from the centre (length of the radius).}$

$\Rightarrow A$

Measurement, STD1 M4 2024 HSC 31

Wombats can run at a speed of 40 km/h over short distances.

At this speed, how many seconds would it take a wombat to run 150 metres? (3 marks)

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$\text{13.5 seconds}$

Show Worked Solution

$\text{Convert km/h to m/sec:}$

$\text{40 km/h}$	$=40\,000\ \text{m/h}$
	$=\dfrac{40\,000}{60 \times 60}\ \text{m/s}$
	$=11.11\ \text{m/s}$

$\therefore\ \text{Time to run 150m}$	$=\dfrac{150}{11.11…}$
	$=13.5\ \text{seconds}$

Financial Maths, STD1 F3 2024 HSC 30

The graph shows the decreasing value of an asset.

For the first 4 years, the value of the asset depreciated by $1500 per year, using a straight-line method of depreciation.

After the end of the 4th year, the method of depreciation changed to the declining-balance method at the rate of 35% per annum.

What is the total depreciation at the end of 10 years? (4 marks)

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$\text{Total depreciation}\ =$46\,681.57$

Show Worked Solution

$\text{Depreciation after 4 years}\ = 4 \times 1500 = $6000$

$\text{Value after 4 years}\ = 50\,000-6000=44\,000$

$\text{Declining balance used for the next 6 years:}$

$V_0=$44\,000, r=0.35, n=6$

$S$	$=V_0(1-r)^n$
	$=44\,000(1-0.35)^6$
	$=$3318.43$

$\therefore\ \text{Total depreciation}\ =50\,000-3318.43=$46\,681.57$

Calculus, 2ADV C4 2024 HSC 10 MC

The diagram shows the graph $y = f(x)$.

The point $Q$ is a horizontal point of inflection.

Let $A(x)= \displaystyle \int_0^x f(t)\,dt$.

How many points of inflection does the graph $y=A(x)$ have?

$2$
$3$
$4$
$5$

Show Answers Only

$B$

Show Worked Solution

$A(x) = \displaystyle \int_0^x f(t)\,dt \ \text{(note this definite integral produces a}\ F(x)\text{)}$

$A^{′}(x) = f(x) $

$A^{″}(x) = f^{′}(x) $

$\text{POI requirements:}\ A^{″}(x) = 0\ \text{and sign (concavity) changes either side.}$

$\text{Inspect graph of}\ f(x)\ \text{to find where}\ \ f^{′}(x)=0\ \ \text{and gradient changes either side of possible points.}$

$\text{→ the three turning points all qualify}$

$\text{→ point}\ Q\ \text{does not qualify (gradient is positive both sides)}$

$\Rightarrow B$

Calculus, 2ADV C3 2024 HSC 31

Two circles have the same centre $O$. The smaller circle has radius 1 cm, while the larger circle has radius $(1 + x)$ cm. The circles enclose a region $QRST$, which is subtended by an angle $\theta$ at $O$, as shaded.

The area of $QRST$ is $A$ cm$^{2}$, where $A$ is a constant and $A \gt 0$.

Let $P$ cm be the perimeter of $QRST$.

By finding expressions for the area and perimeter of $QRST$, show that $P(x)=2x+\dfrac{2A}{x}$. (3 marks)
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Show that if the perimeter, $P(x)$, is minimised, then $\theta$ must be less than 2. (3 marks)
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a. $\text{Area}\ =\ \text{Sector}\ OQR-\text{Sector}\ OTS$

$A$	$=\dfrac{\theta}{2\pi} \times \pi (1+x)^2-\dfrac{\theta}{2\pi} \times \pi \times 1^2$
$A$	$=\dfrac{\theta}{2} (1+x)^2-\dfrac{\theta}{2}$
$A$	$=\dfrac{\theta}{2}(1+2x+x^2-1)$
$A$	$=\dfrac{\theta}{2}x(x+2)$
$\theta$	$=\dfrac{2A}{x(x+2)}$

$\text{Perimeter}\ QRST$	$=\dfrac{\theta}{2\pi} \times 2\pi(x+1)+\dfrac{\theta}{2\pi} \times 2\pi (1)+2x $
	$=\theta(x+1)+\theta+2x$
	$=\theta(x+2)+2x$
	$=\dfrac{2A}{x(x+2)}(x+2)+2x$
	$=2x+\dfrac{2A}{x}$

b. $P(x)=2x+\dfrac{2A}{x}$

$P^{′}(x) = 2-\dfrac{2A}{x^2}$

$P^{″}(x) = \dfrac{4A}{x^3} \gt0\ \ (x,A \gt 0)$

$\Rightarrow\ \text{MIN when}\ \ P^{′}(x)=0:$

$2-\dfrac{2A}{x^2}$	$=0$
$2x^2$	$=2A$
$x^2$	$=A$

$\text{Using part (a):}$

$\theta$	$=\dfrac{2A}{x(x-2)}$
	$=\dfrac{2x^2}{x(x+2)}$
	$=\dfrac{2x}{x+2}$
	$=2 \times \dfrac{x}{x+2} $
	$ \lt 2\ \ \Big(\text{since}\ \ \dfrac{x}{x+2} \lt 1 \ \ \text{for all}\ x \Big) $

$\therefore\ \text{If}\ P(x)\ \text{is minimised,}\ \theta \lt 2. $

Show Worked Solution

a. $\text{Area}\ =\ \text{Sector}\ OQR-\text{Sector}\ OTS$

$A$	$=\dfrac{\theta}{2\pi} \times \pi (1+x)^2-\dfrac{\theta}{2\pi} \times \pi \times 1^2$
$A$	$=\dfrac{\theta}{2} (1+x)^2-\dfrac{\theta}{2}$
$A$	$=\dfrac{\theta}{2}(1+2x+x^2-1)$
$A$	$=\dfrac{\theta}{2}x(x+2)$
$\theta$	$=\dfrac{2A}{x(x+2)}$

$\text{Perimeter}\ QRST$	$=\dfrac{\theta}{2\pi} \times 2\pi(x+1)+\dfrac{\theta}{2\pi} \times 2\pi (1)+2x $
	$=\theta(x+1)+\theta+2x$
	$=\theta(x+2)+2x$
	$=\dfrac{2A}{x(x+2)}(x+2)+2x$
	$=2x+\dfrac{2A}{x}$

COMMENT: Sector/arc calculations used in solution are for those who don’t want to remember formulas.

b. $P(x)=2x+\dfrac{2A}{x}$

$P^{′}(x) = 2-\dfrac{2A}{x^2}$

$P^{″}(x) = \dfrac{4A}{x^3} \gt0\ \ (x,A \gt 0)$

$\Rightarrow\ \text{MIN when}\ \ P^{′}(x)=0:$

$2-\dfrac{2A}{x^2}$	$=0$
$2x^2$	$=2A$
$x^2$	$=A$

$\text{Using part (a):}$

$\theta$	$=\dfrac{2A}{x(x-2)}$
	$=\dfrac{2x^2}{x(x+2)}$
	$=\dfrac{2x}{x+2}$
	$=2 \times \dfrac{x}{x+2} $
	$ \lt 2\ \ \Big(\text{since}\ \ \dfrac{x}{x+2} \lt 1 \ \ \text{for all}\ x \Big) $

$\therefore\ \text{If}\ P(x)\ \text{is minimised,}\ \theta \lt 2. $

Financial Maths, STD2 F5 2024 HSC 41

Twenty-five years ago, Phoenix deposited a single sum of money into a new bank account, earning 2.4% interest per annum compounding monthly.

Present value interest factors for an annuity of $1 for various interest rates $(r)$ and numbers of periods $(n)$ are given in the table.

Phoenix made the following withdrawals from this account.

$2000 at the end of each month for the first 15 years, starting at the end of the first month.
$1200 at the end of each month for the next 10 years, starting at the end of the 181st month after the account was opened.

Calculate the minimum sum that Phoenix could have deposited in order to make these withdrawals. (4 marks)

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$\text{Minimum deposit}\ = $391\,344.80$

Show Worked Solution

$\text{1st Annuity}$

$\text{Find PVA for \$2000 paid monthly for 1st 15 years:}$

$r= \dfrac{2.4%}{12} = 0.2\% = 0.002$

$\text{Total payments (to Phoenix)}\ = 15 \times 12 = 180$

$\text{PVA factor (from table)}\ = 151.036$

$\text{PVA (1st annuity)}\ = 2000 \times 151.036 = $302\,072$

$\text{2nd Annuity}$

$\text{Find PVA for \$1200 paid monthly from year 16 to 25:}$

$\text{PVA (2nd annuity) = PVA (25 years) }-\text{ PVA (15 years)}$

$r= \dfrac{2.4%}{12} = 0.2\% = 0.002$

$\text{Total payments (25 years)}\ = 25 \times 12 = 300$

$\text{PVA factors (from table): 225.430 (25 years), 151.036 (15 years)}$

$\text{PVA (2nd annuity)}$	$=(1200 \times 225.430)-(1200 \times 151.036)$
	$=$89\,272.80$

$\therefore\ \text{Minimum deposit}\ = 302\,072+89\,272.80 = $391\,344.80$

Measurement, STD2 M6 2024 HSC 40

A compass radial survey is shown.

Given that $AC$ is a straight line, find the bearing of $C$ from $O$, correct to the nearest degree. (3 marks)

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$\text{Bearing of}\ C\ \text{from}\ O =220^{\circ}$

Show Worked Solution

$\text{In}\ \Delta ODA,\ \text{using cosine rule:}$

$\cos \angle DOA$	$=\dfrac{38^2+42^2-67.6^2}{2 \times 38 \times 42}$
	$=-0.4223…$
$\angle DOA$	$=115.0^{\circ}$

$\text{Let}\ X\ \text{be a point directly north of point}\ O:$

$\angle DOX=360-285=75^{\circ}$

$\angle XOA=115-75=40^{\circ}$

$\angle AOB=110-40=70^{\circ}$

$\angle COB=180-70=110^{\circ}\ \ (AOC\ \text{is a straight line)}$

$\therefore\ \text{Bearing of}\ C\ \text{from}\ O = 110+110=220^{\circ}$

STRATEGY: The information in $\Delta ODA$ strongly suggests finding $\angle DOA$ is a good strategy to explore.

Statistics, STD2 S5 2024 HSC 35

A random variable is normally distributed with mean 0 and standard deviation 1. The table gives the probability that this random variable is less than $z$.

\begin{array} {|c|c|c|c|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} z \rule[-1ex]{0pt}{0pt} & 0.6 & 0.7 & 0.8 & 0.9 & 1.0 & 1.1 & 1.2 & 1.3 & 1.4 \\
\hline
\rule{0pt}{2.5ex} \textit{Probability} \rule[-1ex]{0pt}{0pt} & 0.7257 & 0.7580 & 0.7881 & 0.8159 & 0.8413 & 0.8643 & 0.8849 & 0.9032 & 0.9192 \\
\hline
\end{array}

The probability values given in the table for different values of $z$ are represented by the shaded area in the following diagram.

The scores in a university examination with a large number of candidates are normally distributed with mean 58 and standard deviation 15.

By calculating a $z$-score, find the percentage of scores that are between 58 and 70. (2 marks)
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Explain why the percentage of scores between 46 and 70 is twice your answer to part (a). (1 mark)
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By using the values in the table above, find an approximate minimum score that a candidate would need to be placed in the top 10% of the candidates. (2 marks)
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a. $28.81%$

b. $\text{Normal distribution is symmetrical about the mean (58).}$

$\text{Since 70 and 46 are both the same distance (12) from the mean, the percentage}$

$\text{of scores in this range will be twice the answer in part (a).}$

c. $\text{Approx minimum score = 78%}$

Show Worked Solution

a. $z\text{-score (58)}\ =\dfrac{x-\mu}{\sigma} = \dfrac{58-58}{15}=0$

$z\text{-score (70)}\ = \dfrac{70-58}{15}=0.8$

$\text{Using table:}$

$\text{% between 58–70}\ =0.7881-0.5=0.2881=28.81%$

b. $\text{Normal distribution is symmetrical about the mean (58).}$

$\text{Since 70 and 46 are both the same distance (12) from the mean, the percentage}$

$\text{of scores in this range will be twice the answer in part (a).}$

c. $z\text{-score 1.3 has a table value 0.9032}$

$1-0.9032=0.0968\ \Rightarrow\ \text{i.e. 9.68% of students score higher.}$

$\text{Find}\ x\ \text{for a}\ z\text{-score of 1.3:}$

$1.3$	$=\dfrac{x-58}{15}$
$x$	$=1.3 \times 15 +58$
	$=77.5$

$\therefore\ \text{Approx minimum score = 78%}$

Statistics, STD2 S1 2024 HSC 15 MC

Some data are used to create a box plot shown.

A histogram is created from the same set of data.

Which of these histograms is NOT possible for the given box plot?

Show Answers Only

$D$

Show Worked Solution

$\text{By inspection of box plot,}\ \ IQR=\dfrac{2}{3}\times \text{range}$

$\text{ln options A, B and C (given 16 data points):}$

$Q_1\ \text{(4th data point)}\ \rightarrow \text{2nd column}$

$Q_3\ \text{(13th data point)}\ \rightarrow \text{6th column}$

$\text{Since}\ IQR=\dfrac{2}{3}\times \text{range}\ \Rightarrow\ \text{histograms are possible.}$

$\text{ln option D:}$

$Q_1 \rightarrow \text{3rd column}$

$Q_3 \rightarrow \text{5th column}$

$\text{Since}\ IQR=\dfrac{1}{3} \times \text{range}\ \Rightarrow\ \text{not possible.}$

$\Rightarrow D$

Financial Maths, 2ADV M1 2024 HSC 30

Suppose the geometric series $x+x^2+x^3+\ \cdots$ has a limiting sum.

By considering the graph $y=-1-\dfrac{1}{x-1}$, or otherwise, find the range of possible values of $S$. (3 marks)

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$S \in \Big( -\dfrac{1}{2}, \infty\Big ) $

Show Worked Solution

$x + x^2+x^3\ \cdots \Rightarrow a=x, r=x$

$S=\dfrac{a}{1-r} = \dfrac{x}{1-x}=\dfrac{-(1-x)+1}{1-x} = -1+\dfrac{1}{1-x}=-1-\dfrac{1}{x-1}$

$\text{If limiting sum}\ \Rightarrow \abs{r} \lt 1\ \Rightarrow \ \abs{x} \lt 1$

$\Rightarrow \text{Possible values of}\ S =\ \text{range of}\ \ y=-1-\dfrac{1}{x-1}\ \text{in domain}\ \abs{x} \lt 1$

$\text{Consider}\ \ y=-1-\dfrac{1}{x-1}:$

$\text{Asymptote at}\ \ x=1.$

\begin{array} {|c|c|c|c|c|c|}
\hline
\rule{0pt}{2.5ex} x \rule[-1ex]{0pt}{0pt} & -2 & -1 & 0 & 1 & 2 \\
\hline
\rule{0pt}{2.5ex} y \rule[-1ex]{0pt}{0pt} & -\dfrac{2}{3} & -\dfrac{1}{2} & \ \ 0\ \ & \infty & -2 \\
\hline
\end{array}

$\text{As}\ x \rightarrow \infty, \ y \rightarrow -1$

$\text{As}\ x \rightarrow -\infty, \ y \rightarrow -1$

$\abs{x} \lt 1\ \Rightarrow \ y \in \Big( -\dfrac{1}{2}, \infty\Big ) $

$\therefore\ \text{Possible values of}\ S \in \Big( -\dfrac{1}{2}, \infty\Big ) $

CHEMISTRY, M2 EQ-Bank 8

A gas at a temperature of $9.0 \times 10^2\ \text{K}$ in a container with a volume of $30.0\ \text{L}$ has a pressure of $5.0 \times 10^2\ \text{kPa}$.

How many moles of gas are in the container? (3 marks)

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If the empty container weighs $450.0\ \text{g}$ and the container with the gas weighs $526.0\ \text{g}$, what is the gaseous element in the container? (2 marks)

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a. $2.0\ \text{mol}$

b. The unknown gas is fluorine.

Show Worked Solution

a. Using the Ideal Gas Law:

$PV=nRT \Rightarrow n=\dfrac{PV}{RT}$

$n=\dfrac{500 \times 30}{8.314 \times 900}=2.0\ \text{mol (2 sig.fig)}$

b. Mass of the gas $=526.0-450.0=76\ \text{g}$

Molar Mass of the gas $=\dfrac{m}{n}=\dfrac{76}{2.0}=38\ \text{g mol}^{-1}$

→ This is equal to the MM of fluorine gas $\ce{(F2)}$.

→ The unknown gas is fluorine.

Measurement, STD2 M1 2024 HSC 34

A container for soccer balls is made using two half spheres joined to each end of a cylindrical body.

Three soccer balls fit exactly inside the container. Each ball has a diameter of 23 cm.

The hemispherical ends of the container just touch the surface of the soccer balls.

What is the total surface area of the container? Give your answer in square metres correct to one decimal place. (4 marks)

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$\text{0.5 m}^{2}$

Show Worked Solution

$\text{Ball diameter = 23 cm}\ \Rightarrow \ r=11.5\ \text{cm}$

$\text{S.A. (2 half spheres)}$	$=2 \times \dfrac{1}{2} \times 4 \pi \times 11.5^2$
	$=1662\ \text{cm}^{2}$

$\text{Cylinder width}\ =\ \text{ball circumference}= 2 \times \pi \times 11.5$

$\text{Cylinder height}\ = 2 \times 23 = 46\ \text{cm}$

$\text{S.A. (cylinder)}\ = 2\pi rh = 2 \times \pi \times 11.5 \times 46 = 3324\ \text{cm}^{2}$

$\text{Note:}\ 1\ \text{m}^{2} = 100\ \text{cm} \times 100\ \text{cm}\ = 10\,000\ \text{cm}^{2}$

$\text{S.A. (container)}$	$=1662 + 3324$
	$=4986\ \text{cm}^{2}$
	$=\dfrac{4986}{10\,000}\ \text{m}^{2}$
	$=0.5\ \text{m}^{2}\ \text{(1 d.p.)}$

Measurement, STD2 M7 2024 HSC 33

Wombats can run at a speed of 40 km/h over short distances.

At this speed, how many seconds would it take a wombat to run 150 metres? (3 marks)

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$\text{13.5 seconds}$

Show Worked Solution

$\text{Convert km/h to m/sec:}$

$\text{40 km/h}$	$=40\,000\ \text{m/h}$
	$=\dfrac{40\,000}{60 \times 60}\ \text{m/s}$
	$=11.11\ \text{m/s}$

$\therefore\ \text{Time to run 150m}$	$=\dfrac{150}{11.11…}$
	$=13.5\ \text{seconds}$

CHEMISTRY, M3 EQ-Bank 8

A galvanic cell has been set up as illustrated in the diagram below.

The standard potential for this reaction is 0.78 V. Use half equations to determine the unknown electrode. (2 marks)

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The unknown solution is light green in colour. Explain what will happen to the colour of the unknown solution as the reaction proceeds. (2 marks)

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After some time, a solid deposit formed on the copper electrode was removed and dried. The mass of the deposit was 0.150 g. Determine the final concentration of the copper nitrate solution. (3 marks)

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a. $\ce{Fe(s) -> Fe^{2+} + 2e^-}$

b. As the reaction progresses:

→ The solution will darken, taking on a more intense green colour.

→ This change occurs because nickel undergoes oxidation to form $\ce{Fe^{2+}}$ ions, which are released into the solution, thereby increasing its colour intensity.

c. $0.137\ \text{mol L}^{-1}$

Show Worked Solution

a.	$E^{\circ}_{\text{cell}}$	$=E^{\circ}_{\text{cathode}}-E^{\circ}_{\text{anode}}$
	$0.78$	$=0.34-E^{\circ}_{\text{anode}}$
	$E^{\circ}_{\text{anode}}$	$=0.34-0.78$
	$E^{\circ}_{\text{anode}}$	$=-0.44$

→ $\ce{Fe^{2+} + 2e^- -> Fe(s)} \qquad -0.44\ \text{V}$

→ $\ce{Fe^{2+}}$ is undergoing oxidation. The correct half equation is: $\ce{Fe(s) -> Fe^{2+} + 2e^-}$

b. As the reaction progresses:

→ The solution will darken, taking on a more intense green colour.

→ This change occurs because nickel undergoes oxidation to form $\ce{Fe^{2+}}$ ions, which are released into the solution, thereby increasing its colour intensity.

c. Moles of solid copper formed on electrode: $\dfrac{m}{MM}=\dfrac{0.175}{63.55}=2.36 \times 10^{-3}\ \text{mol}$

Moles of copper taken out of solution: $2.36 \times 10^{-3}\ \text{mol}$

Moles of copper remaining in solution: $(0.15 \times 0.18)-2.36 \times 10^{-3}= 0.0246\ \text{mol}$

Final concentration: $c=\dfrac{n}{V}=\dfrac{0.0246}{0.18}=0.137\ \text{mol L}^{-1}$

CHEMISTRY, M3 EQ-Bank 20

During a laboratory investigation, a student mixed two solutions and observed a sudden colour change, an increase in temperature, and the formation of bubbles.

Explain why these observations indicate a chemical change. (3 marks)

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Discuss two types of chemical reactions that could cause at least two of these observations each. (2 marks)

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a. Colour Change:

→ This suggests that new chemical compounds are forming that have different properties from the original reactants.

Temperature Increase (Exothermic Reaction):

→ The release of heat indicates that the reaction is exothermic, where energy is released as bonds are formed in the products.

Gas Production (Bubbles):

→ The formation of bubbles without boiling is a sign that a gas is being produced as a result of the reaction.

b. Acid-Base Reaction:

→ When an acid reacts with a base, it can lead to a colour change if an indicator is present (e.g., phenolphthalein changes from colourless to pink). An exothermic reaction may also occur, causing a temperature increase.

Decomposition Reaction:

→ Certain decomposition reactions, such as the breakdown of hydrogen peroxide, produce gas (oxygen) and heat. The bubbling and increase in temperature can be observed in this type of reaction.

Show Worked Solution

a. Colour Change:

→ This suggests that new chemical compounds are forming that have different properties from the original reactants.

Temperature Increase (Exothermic Reaction):

→ The release of heat indicates that the reaction is exothermic, where energy is released as bonds are formed in the products.

Gas Production (Bubbles):

→ The formation of bubbles without boiling is a sign that a gas is being produced as a result of the reaction.

b. Acid-Base Reaction:

Decomposition Reaction:

→ Certain decomposition reactions, such as the breakdown of hydrogen peroxide, produce gas (oxygen) and heat. The bubbling and increase in temperature can be observed in this type of reaction.

CHEMISTRY, M3 EQ-Bank 16

Describe how activation energy, collision frequency, and molecular orientation work together to determine the rate of a chemical reaction. In your answer, define what each term refers to and relate these factors to collision theory. (5 marks)

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→ Activation Energy: For a reaction to occur, the colliding molecules must have enough energy to overcome the activation energy barrier, which is the minimum energy required to break the bonds in the reactants and initiate the reaction.

→ Collision Frequency: The rate of a reaction is also influenced by how frequently reactant molecules collide.

→ Molecular Orientation: In addition to having enough energy, molecules must collide with the correct orientation for a reaction to take place. Reactant molecules need to align in a way that allows bonds to break and new bonds to form.

→ Increasing collision frequency increases the number of opportunities for molecules to collide, but only those collisions with enough energy and the correct orientation will lead to successful bond rearrangements.

→ For the maximum rate of reaction there needs to be a lower activation energy which makes it easier for collisions to result in a reaction, the proper orientation that ensures when collisions occur, they lead to the formation of products and a high collision frequency.

Show Worked Solution

→ Collision Frequency: The rate of a reaction is also influenced by how frequently reactant molecules collide.

CHEMISTRY, M3 EQ-Bank 28v4

A student stirs 2.50 g of silver (I) nitrate powder into 100.0 mL of 1.50 mol L$^{-1}$ sodium chloride solution until it is fully dissolved. A reaction occurs and a precipitate appears.

Write a balanced chemical equation for the reaction. (1 mark)

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Calculate the theoretical mass of precipitate that will be formed. (4 marks)

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The student weighed a piece of filter paper, filtered out the precipitate and dried it thoroughly in an incubator. The final precipitate mass was higher than predicted in (b).
Identify one scientific reason why the precipitate mass was too high and suggest an improvement to the experimental method which would eliminate this error. (2 marks)

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a. $\ce{AgNO3(aq) + NaCl(aq) -> AgCl(s) + NaNO3(aq)}$

b. $2.11 \text{ g}$

c. One possible reason for the higher mass could be the presence of excess solution in the filter paper or incomplete drying. An improvement could be to ensure the precipitate is thoroughly rinsed with distilled water to remove soluble impurities and dried completely in the incubator before weighing.

Show Worked Solution

a. $\ce{AgNO3(aq) + NaCl(aq) -> AgCl(s) + NaNO3(aq)}$

b. $\ce{n(AgNO3) = \frac{m}{M} = \frac{2.50}{107.9 + 14.01 + 16.00 \times 3} = \frac{2.50}{169.91} = 0.01471 \text{ mol}}$

$\ce{n(NaCl) = c \times V = 1.50 \times 0.100 = 0.150 \text{ mol}}$

$\ce{AgNO3} \text{ is the limiting reagent}$

$\ce{n(AgCl) = n(AgNO3) = 0.01471 \text{ mol}}$

$\ce{m(AgCl) = n \times M = 0.01471 \times (107.9 + 35.45) = 0.01471 \times 143.35 = 2.11 \text{ g}}$

CHEMISTRY, M3 EQ-Bank 28v2

A student stirs 3.50 g of copper (II) nitrate powder into 150.0 mL of 1.50 mol L$^{-1}$ sodium chloride solution until it is fully dissolved. A reaction occurs and a precipitate appears.

Write a balanced chemical equation for the reaction. (1 mark)

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Calculate the theoretical mass of precipitate that will be formed. (4 marks)

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The student weighed a piece of filter paper, filtered out the precipitate and dried it thoroughly in an incubator. The final precipitate mass was higher than predicted in (b).
Identify one scientific reason why the precipitate mass was too high and suggest an improvement to the experimental method which would eliminate this error. (2 marks)

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a. $\ce{Cu(NO3)2(aq) + 2NaCl(aq) -> CuCl2(s) + 2NaNO3(aq)}$

b. $2.51 \text{ g}$

Show Worked Solution

a. $\ce{Cu(NO3)2(aq) + 2NaCl(aq) -> CuCl2(s) + 2NaNO3(aq)}$

b. $\ce{n(Cu(NO3)2) = \frac{m}{M} = \frac{3.50}{63.55 + 2 \times (14.01 + 16.00 \times 3)} = \frac{3.50}{187.57} = 0.01866 \text{ mol}}$

$\ce{n(NaCl) = c \times V = 1.50 \times 0.150 = 0.225 \text{ mol}}$

$\ce{Cu(NO3)2} \text{ is the limiting reagent}$

$\ce{n(CuCl2) = n(Cu(NO3)2) = 0.01866 \text{ mol}}$

$\ce{m(CuCl2) = n \times M = 0.01866 \times (63.55 + 2 \times 35.45) = 0.01866 \times 134.45 = 2.51 \text{ g}}$

CHEMISTRY, M4 EQ-Bank 6

Explain why using bond energies is not an accurate method of calculating enthalpy changes. (2 marks)

Show Answers Only

→ Enthalpy changes calculated using bond energies are not entirely accurate because bond energies represent averages of similar bond dissociations.

→ The actual energy of a specific bond can vary depending on the electrochemical environment around it, such as the presence of nearby electronegative atoms, adjacent double bonds, or the overall size of the molecule.

→ The energy of a specific bond depends also on the pressure, temperature, and state of the molecule. Bond energies assume standard laboratory conditions and all compounds already being in gas form and so are inaccurate.

Show Worked Solution

→ Enthalpy changes calculated using bond energies are not entirely accurate because bond energies represent averages of similar bond dissociations.

CHEMISTRY, M4 EQ-Bank 5

The chemical equation for the complete combustion of ethane $\ce{(C2H6)}$ is given below:

$\ce{2C2H6(g) + 7O2(g) -> 4CO2(g) + 6H2O(l)}$

The structural formula for ethane and standard bond energies for provided for you.

\begin{array} {|c|c|}
\hline Bond & Enthalpy \text{ (kJ mol}^{-1}) \\
\hline \ce{C-C} & 348 \\
\hline \ce{O-O} & 146 \\
\hline \ce{O=O} & 495 \\
\hline \ce{C-O} & 358 \\
\hline \ce{C=O} & 799 \\
\hline \ce{C-H} & 413 \\
\hline \ce{H-O} & 463 \\
\hline \end{array}

Using the bond energies provided, calculate the enthalpy for the complete combustion of one mole of ethane. (3 marks)

Show Answers Only

$-1415.5\ \text{kJ mol}^{-1}$

Show Worked Solution

$\Delta H$	$=\Sigma\,{\text{bonds broken}}-\Sigma\,{\text{bonds formed}}$
	$=((12 \times 413) + (2 \times 348) + (7 \times 495))-((8 \times 799) + (12 \times 463))$
	$=9117-11948$
	$=-2831$ (for two moles of ethane, as per the equation)

→ $\Delta H$ for the combustion of one mole of ethane is $-1415.5 \text{ kJ mol}^{-1}$

CHEMISTRY, M4 EQ-Bank 4

The chemical equation for the combustion of butane $\ce{(C4H10)}$ is given below:

$\ce{2C4H10(g) + 13O2(g) -> 8CO2(g) + 10H2O(g)} \qquad \Delta H = -5754\ \text{kJ mol}^{-1}$

Given that the standard enthalpy of formation of $\ce{CO2(g)}$ is –393 kJ mol$^{-1}$ and $\ce{H2O(g)}$ is –241 kJ mol $^{-1}$, calculate the standard enthalpy of formation of butane. (3 marks)

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$-126\ \text{kJ mol}^{-1}$

Show Worked Solution

→ The enthalpy change for the combustion of 2 moles of butane is -5754 kJ.

$\Delta H$	$=\Sigma{\Delta H_f \text{ (products)}}-\Sigma{\Delta H_f \text{ (reactants)}}$
$-5754$	$=(8 \times -393 + 10 \times -241)-(2 \times \Delta H_f \text{ (butane)})$
$-5754$	$=-4334-2 \times \Delta H_f \text{ (butane)}$
$\Delta H_f \text{ (butane)}$	$=\dfrac{-4334 + 5754}{2}$
	$=-126\ \text{kJ mol}^{-1}$

→ The standard enthalpy of formation of butane is $-126\ \text{kJ mol}^{-1}$.

CHEMISTRY, M4 EQ-Bank 1

The chemical equation for the combustion of butanol $\ce{(C4H9OH(l))}$ is given below

$\ce{C4H9OH(l) + 6O2(g) -> 4CO2(g) + 5H2O(l)}$ $\Delta H = -2670\ \text{kJ mol}^{-1}$

\begin{array} {|c|c|}
\hline \text{Compound} & \Delta H_f \ \text{(kJ mol}^{-1}) \\
\hline \ce{CO2(g)} & -393 \\
\hline \ce{H2O(l)} & -286 \\
\hline \end{array}

Define what the term 'standard enthalpy of formation' means. (1 mark)

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Use the table data to calculate the standard enthalpy of formation of butanol. (3 marks)

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a. Standard enthalpy of formation:

→ The change in enthalpy when one mole of a substance is formed from its constituent elements in their standard states under standard conditions (298 K temperature and 100 kPa).

→ The elements must be in their most stable form at these conditions.

b. $-332\ \text{kJ mol}^{-1}$

Show Worked Solution

a. Standard enthalpy of formation:

→ The change in enthalpy when one mole of a substance is formed from its constituent elements in their standard states under standard conditions (298 K temperature and 100 kPa).

→ The elements must be in their most stable form at these conditions.

b.	$\Delta H$	$= \Sigma{\Delta H_f \text{ (products)}}-\Sigma{\Delta H_f \text{ (reactants)}}$
	$-2670$	$=(4 \times -393 + (5 \times -286))-(\Delta H_f \text{ butanol} + (6 \times 0))$
	$\Delta H_f \text{ butanol}$	$=-3002 + 2670$
		$=-332\ \text{kJ mol}^{-1}$

→ The standard enthalpy of formation of an element is 0.

CHEMISTRY, M4 EQ-Bank 17

The decomposition of a metal carbonate is represented by the following equation:

$\ce{MCO3(s) → MO(s) + CO2(g)}$

The following data was recorded:

$\Delta H = +130 \, \text{kJ/mol},\ \ \Delta S = +160 \, \text{J/mol K}$

Calculate the Gibbs free energy at 350 K. (2 marks)

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Determine if the reaction is spontaneous at this temperature. (1 mark)

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Discuss how both enthalpy and entropy influence the spontaneity of this reaction and predict the temperature range in which the reaction will be spontaneous. (4 marks)

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a. $\Delta G = +74 \, \text{kJ/mol}$

b. The reaction is non-spontaneous at 350 K.

c. Enthalpy and entropy influence the spontaneity:

→ To be spontaneous, $\Delta G$ must be negative, where $\Delta G = \Delta H- T\Delta S$.

→ The reaction has a positive enthalpy, indicating it requires energy input, which opposes spontaneity.

→ However, the positive entropy means the disorder of the system increases, which favours spontaneity.

→ At higher temperatures, the entropy contribution will dominate, overcoming the positive enthalpy and making the reaction spontaneous.

→ To find the temperature where the reaction becomes spontaneous, set $\Delta G = 0$:

$0=\Delta H-T\Delta S$

$T=\dfrac{\Delta H}{\Delta S}=\dfrac{130}{0.160}=812.5\ \text{K}$

→ Thus, the reaction will be spontaneous above 812.5 K.

Show Worked Solution

a. $\Delta G = \Delta H- T\Delta S$

Convert $\Delta S$ to $\text{kJ/mol K}$

$\Delta S = 0.160 \, \text{kJ/mol K}$

$\Delta G = 130-(350 \times 0.160) = 130-56 = +74 \, \text{kJ/mol}$

b. Since $\Delta G > 0$, the reaction is non-spontaneous at 350 K.

c. Enthalpy and entropy influence the spontaneity:

→ To be spontaneous, $\Delta G$ must be negative, where $\Delta G = \Delta H- T\Delta S$.

→ The reaction has a positive enthalpy, indicating it requires energy input, which opposes spontaneity.

→ However, the positive entropy means the disorder of the system increases, which favours spontaneity.

→ At higher temperatures, the entropy contribution will dominate, overcoming the positive enthalpy and making the reaction spontaneous.

→ To find the temperature where the reaction becomes spontaneous, set $\Delta G = 0$:

$0=\Delta H-T\Delta S$

$T=\dfrac{\Delta H}{\Delta S}=\dfrac{130}{0.160}=812.5\ \text{K}$

→ Thus, the reaction will be spontaneous above 812.5 K.

v1 Algebra, STD2 A2 2009 HSC 24d

A factory makes both cloth and leather lounges. In any week

• the total number of cloth lounges and leather lounges that are made is 400
• the maximum number of leather lounges made is 270
• the maximum number of cloth lounges made is 325.

The factory manager has drawn a graph to show the numbers of leather lounges ($x$) and cloth lounges ($y$) that can be made.

Find the equation of the line $AD$. (1 mark)

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Explain why this line is only relevant between $B$ and $C$ for this factory. (1 mark)

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The profit per week, $$P$, can be found by using the equation $P = 2520x + 1570y$.

Compare the profits at $B$ and $C$. (2 marks)

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Show Answers Only

$x+y=400$
$\text{Since the max amount of leather lounges}=270$

$\rightarrow\ x\ \text{cannot be}\ >270$

$\text{Since the max amount of cloth lounges}=325$

$\rightarrow\ y\ \text{cannot be}\ >325$

$\therefore\ \text{The line}\ AD\ \text{is only possible between}\ B\ \text{and}\ C.$
$\text{The profits at}\ C\ \text{are }$185\ 250\ \text{more than at}\ B.$

Show Worked Solution

i. $\text{We are told the number of leather lounges}\ (x),$

$\text{and cloth lounges}\ (y),\ \text{made in any week} = 400$

$\rightarrow\ \text{Equation of}\ AD\ \text{is}\ x+y=400$

♦♦♦ Mean mark part (i) 14%.
Using $y=mx+c$ is a less efficient but equally valid method, using $m=–1$ and $b=400$ ($y$-intercept).

ii. $\text{Since the max amount of leather lounges}=270$

$\rightarrow\ x\ \text{cannot}\ >270$

$\text{Since the max amount of cloth lounges}=325$

$\rightarrow\ y\ \text{cannot}\ >325$

$\therefore\ \text{The line}\ AD\ \text{is only possible between}\ B\ \text{and}\ C.$

♦ Mean mark part (ii) 49%.

iii. $\text{At}\ B,\ x=75,\ y=325$

$\rightarrow\ $P (\text{at}\ B)$	$=2520\times 75+1570\times 325$
	$=189\ 000+510\ 250$
	$=$699\ 250$

$\text{At}\ C,\ x=270,\ y=130$

$\rightarrow\ $P (\text{at}\ C)$	$=2520\times 270+1570\times 130$
	$=680\ 400+204\ 100$
	$=$884\ 500$

$\text{Difference in profits}=$884\ 500-$699\ 250=$185\ 250$

$\text{The profits at}\ C\ \text{are } $185\ 250\ \text{more than at}\ B.$

♦ Mean mark (iii)40%.

v1 Algebra, STD2 A2 2010 HSC 27c

The graph shows tax payable against taxable income, in thousands of dollars.

Use the graph to find the tax payable on a taxable income of $$18\ 000$. (1 mark)

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Use suitable points from the graph to show that the gradient of the section of the graph marked $A$ is $\dfrac{7}{15}$. (1 mark)

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How much of each dollar earned between $$18\ 000$ and $$33\ 000$ is payable in tax? Give your answer correct to the nearest whole number. (1 mark)

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Write an equation that could be used to calculate the tax payable, $T$, in terms of the taxable income, $I$, for taxable incomes between $$18\ 000$ and $$33\ 000$. (2 marks)

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Show Answers Only

$$3000\ \ \text{(from graph)}$
$\text{See worked solution}$
$46\frac{2}{3}\approx 47\ \text{cents per dollar earned}$
$\text{Tax payable →}\ T=\dfrac{7}{15}I-5400$

Show Worked Solution

$\text{Income on}\ $18\ 000=$3000\ \ \text{(from graph)}$

ii. $\text{Using the points}\ (18, 3)\ \text{and}\ (33, 10)$

$\text{Gradient at}\ A$	$=\dfrac{y_2-y_1}{x_2-x_1}$
	$=\dfrac{10\ 000-3000}{33\ 000-18\ 000}$
	$=\dfrac{7000}{15\ 000}$
	$=\dfrac{7}{15}\ \ \ \ \text{… as required}$

♦♦ Mean mark (ii) 25%.

iii. $\text{The gradient represents the tax applicable on each dollar}$

$\text{Tax}$	$=\dfrac{7}{15}\ \text{of each dollar earned}$
	$=46\frac{2}{3}\approx 47\ \text{cents per dollar earned (nearest whole number)}$

♦♦♦ Mean mark (iii) 12%!
MARKER’S COMMENT: Interpreting gradients is an examiner favourite, so make sure you are confident in this area.

iv. $\text{Tax payable up to }$18\ 000 = $3000$

$\text{Tax payable on income between }$18\ 000\ \text{and }$33\ 000$

$=\dfrac{7}{15}(I-18\ 000)$

$\therefore\ \text{Tax payable →}\ \ T$	$=3000+\dfrac{7}{15}(I-18\ 000)$
	$=3000+\dfrac{7}{15} I-8400$
	$=\dfrac{7}{15}I-5400$

♦♦♦ Mean mark (iv) 15%.
STRATEGY: The earlier parts of this question direct students to the most efficient way to solve this question. Make sure earlier parts of a question are front and centre of your mind when devising strategy.

v1 Algebra, STD2 A2 2007 HSC 18 MC

Art started to make this pattern of shapes using matchsticks.

If the pattern of shapes is continued, which shape would use exactly 416 matchsticks?

Shape 83
Shape 103
Shape 104
Shape 138

Show Answers Only

$D$

Show Worked Solution

\begin{array} {|l|c|c|c|}
\hline
\rule{0pt}{2.5ex} \text{Shape}\ \textit(S) \rule[-1ex]{0pt}{0pt}\ \ &\ \ 1\ \ &\ \ 2\ \ &\ \ 3\ \ \\
\hline
\rule{0pt}{2.5ex} \text{Matches}\ \textit(M) \rule[-1ex]{0pt}{0pt} \ \ & \ \ 5\ \ &\ \ 8\ \ &\ \ 11\ \ \\
\hline
\end{array}

$\text{Equation rule:}$

$M=3S+2$

$\text{Find}\ \ S\ \text{when}\ \ M=416:$

$416$	$=3S+2$
$3S$	$=414$
$S$	$=138$

$\therefore\ \text{The 138th shape uses 416 matchsticks.}$

$\Rightarrow D$

v1 Algebra, STD2 A1 2020 HSC 13 MC

When Stuart stops drinking alcohol at 11:30 pm, he has a blood alcohol content (BAC) of 0.08625.

The number of hours required for a person to reach zero BAC after they stop consuming alcohol is given by the formula:

$\text{Time}=\dfrac{BAC}{0.015}$.

At what time on the next day should Stuart expect his BAC to be 0.05?

1:33 am
1:55 am
2:15 am
5:15 am

Show Answers Only

$B$

Show Worked Solution

$\text{Time from 0.08625 → 0}\ BAC$

$=\dfrac{0.08625}{0.015}$

$=5.75\ \text{hours}$

$\text{Time from 0.08625 → 0.05}\ BAC$

$=\dfrac{(0.08625 – 0.05)}{0.08625}\times 5.75$

$=\dfrac{29}{69}\times 5.75$

$=2.41\dot{6}=2\ \text{h}\ 25\ \text{min}$

$\therefore\ \text{Time}$	$=11:30\ \text{pm} \ + 2 \ \text{h} \ 25 \ \text{min}$
	$=1:55\ \text{am}$

$\Rightarrow B$

♦♦♦ Mean mark 17%.
COMMENT: The rates aspect of this question proved extremely challenging.

v1 Algebra, STD2 A1 2007 HSC 28b

This shape is made up of a right-angled triangle and a regular hexagon.

The area of a regular hexagon can be estimated using the formula $A=2.598H^2$ where $H$ is the side-length.

Calculate the total area of the shape using this formula. (3 marks)

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Show Answers Only

$51.264\ \text{cm}^2$

Show Worked Solution

$\text{Area}=2.598H^2$

$\text{Using Pythagoras}$

$H^2$	$= 3^2+3^2$
	$=18$
$H$	$=\sqrt{18}$

$\therefore\ \text{Area of hexagon}$	$=2.598\times {\sqrt {18}}\ ^2$
	$=46.764\ \text{cm}^2$

$\text{Area of triangle}$	$=\dfrac{1}{2}bh$
	$=\dfrac{1}{2}\times 3\times 3$
	$=4.5\ \text{cm}^2$

$\therefore\ \text{Total Area}$	$=46.764+4.5$
	$=51.264\ \text{cm}^2$

EXAMCOPY Functions, MET2 2022 VCAA 4

Consider the function `f`, where `f:\left(-\frac{1}{2}, \frac{1}{2}\right) \rightarrow R, f(x)=\log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right).`

Part of the graph of `y=f(x)` is shown below.

State the range of `f(x)`. (1 mark)

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i. Find `f^{\prime}(0)`. (2 marks)

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ii. State the maximal domain over which `f` is strictly increasing. (1 mark)

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Show that `f(x)+f(-x)=0`. (1 mark)

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Find the domain and the rule of `f^{-1}`, the inverse of `f`. (3 marks)

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Let `h` be the function `h:\left(-\frac{1}{2}, \frac{1}{2}\right) \rightarrow R, h(x)=\frac{1}{k}\left(\log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right)\right)`, where `k \in R` and `k>0`.
The inverse function of `h` is defined by `h^{-1}: R \rightarrow R, h^{-1}(x)=\frac{e^{k x}-1}{2\left(e^{k x}+1\right)}`.
The area of the regions bound by the functions `h` and `h^{-1}` can be expressed as a function, `A(k)`.
The graph below shows the relevant area shaded.

You are not required to find or define `A(k)`.

Determine the range of values of `k` such that `A(k)>0`. (1 mark)

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Explain why the domain of `A(k)` does not include all values of `k`. (1 mark)

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a.	`R`
b.i	`f^{\prime}(0)=4`
b.ii	`\left(-\frac{1}{2}, \frac{1}{2}\right)`
c.	`0`
d.	`x \in \mathbb{R}`
e.i	` k > 4`
e.ii	No bounded area for `0<k \leq 4`

Show Worked Solution

a. `R` is the range.

b.i	`f(x)`	`= \log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right)`
	`f^{\prime}(x)`	`= \frac{1}{x+\frac{1}{2}}+\frac{1}{\frac{1}{2}-x}`
		`= \frac{2}{2 x+1}-\frac{2}{2 x-1}`
	`f^{\prime}(0)`	`= \frac{2}{2 xx 0+1}-\frac{2}{2 xx 0-1}`
		`= 4`

b.ii `\left(-\frac{1}{2}, \frac{1}{2}\right)`

c. `f(x)+f(-x)`	`= \log _e\left(x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-x\right)+\log _e\left(-x+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}+x\right)`
	`= 0`

d. To find the inverse swap `x` and `y` in `y=f(x)`

`x`	`= \log _e\left(y+\frac{1}{2}\right)-\log _e\left(\frac{1}{2}-y\right)`
`x`	`= \log _e\left(\frac{y+\frac{1}{2}}{\frac{1}{2}-y}\right)`
`e^x`	`=\frac{y+\frac{1}{2}}{\frac{1}{2}-y}`
`y+\frac{1}{2}`	`= e^x\left(-y+\frac{1}{2}\right)`
`y+\frac{1}{2}`	`= -e^x y+\frac{e^x}{2}`
`y\left(e^x+1\right)`	`= \frac{e^x-1}{2}`
`:.\ f^(-1)(x)`	`= \frac{e^x-1}{2(e^x + 1)}`

`:.` Domain: `x \in \mathbb{R}`

e.i The vertical dilation factor of `f(x)` is `1/k`

For `A(k)>=0` , `h^{\prime}(0)<1`

`\frac{1}{k}(4)<1` [Using CAS]

`:.\ k > 4`

♦♦♦♦ Mean mark (e.i) 10%.
MARKER’S COMMENT: Incorrect responses included `k>0` and `4<k<33`.

e.ii When `h \geq h^{-1}` for `x>0` (or `h \leq h^{-1}` for `x<0`) there is no bounded area.

`:.` There will be no bounded area for `0<k \leq 4`.

♦♦♦♦ Mean mark (e.ii) 10%.

CHEMISTRY, M8 2022 VCE 5*

A chemist uses spectroscopy to identify an unknown organic molecule, Molecule $\text{J}$, that contains chlorine.

The ${}^{13}\text{C NMR}$ spectrum of Molecule $\text{J}$ is shown below.

The infra-red (IR) spectrum of Molecule $\text{J}$ is shown below.

Name the functional group that produces the peak at 168 ppm in the ${}^{13}\text{C NMR}$ spectrum on the first image, which is consistent with the IR spectrum shown above. Justify your answer with reference to the IR spectrum. (2 marks)

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The mass spectrum of Molecule $\text{J}$ is shown below

The molecular mass of Molecule $\text{J}$ is 108.5
Explain the presence of the peak at 110 m/z. (1 mark)

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The ${ }^1 \text{H NMR}$ spectrum of Molecule $\text{J}$ is shown below.

The ${ }^1 \text{H NMR}$ spectrum consists of two singlet peaks.
What information does this give about the molecule? (2 marks)

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Draw a structural formula for Molecule $\text{J}$ that is consistent with the information provided in parts a–c. (2 marks)

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a. → Absence of a very broad $\ce{OH}$ acid peak between 2500–3000.

→ Molecule $\text{J}$ must be an ester.

b. The peak at 110 m/z:

→ due to the Chlorine-37 isotope which is slightly heavier than the more abundant Chlorine-35.

c. The two singlet peaks indicate:

→ two different hydrogen environments within the molecule.

→ there are no adjacent hydrogen environments.

→ The relative heights of the peaks show the ratios of the hydrogens in the environments are 2 : 3.

d. Either of the two molecules show below are correct:

Show Worked Solution

a. → Absence of a very broad $\ce{OH}$ acid peak between 2500–3000.

→ Molecule $\text{J}$ must be an ester.

♦ Mean mark (a) 41%.

b. The peak at 110 m/z:

→ due to the Chlorine-37 isotope which is slightly heavier than the more abundant Chlorine-35.

c. The two singlet peaks indicate:

→ two different hydrogen environments within the molecule.

→ there are no adjacent hydrogen environments.

→ The relative heights of the peaks show the ratios of the hydrogens in the environments are 2 : 3.

♦♦♦ Mean mark (b) 15%.
COMMENT: Know the masses of common isotopes.

d. Either of the two molecules show below are correct:

♦ Mean mark (d) 40%.

CHEMISTRY, M8 2021 VCE 16 MC

Which one of the following statements about IR spectroscopy is correct?

IR radiation changes the spin state of electrons.
Bond wave number is influenced only by bond strength.
An IR spectrum can be used to determine the purity of a sample.
In an IR spectrum, high transmittance corresponds to high absorption.

Show Answers Only

$C$

Show Worked Solution

→ Every pure compound has a different fingerprint region on the Infrared spectrum.

→ Hence the fingerprint region of the sample can be compared against the fingerprint region of the pure substance to determine the purity of the sample.

$\Rightarrow C$

♦♦♦ Mean mark 15%.

CHEMISTRY, M8 2023 VCE 7-2*

The infrared (IR) spectrum of the molecule 3-methyl-2-butanone is shown below.

Explain why different frequencies of infrared radiation can be absorbed by the same molecule as shown in the spectrum above. (3 marks)

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Show Answers Only

→ As infrared radiation is passed through the molecules, the different bonds within the molecule vibrate at specific wavelengths leading to the absorption of the infrared radiation.

→ In this way, different frequencies of infrared radiation can be absorbed by a molecule as bonds differ in electronegativity, dipole strengths and in the masses of atoms at the end of bonds.

→ For example, the double bond between the oxygen atom and carbon atom in the given molecule has a greater dipole than the carbon-hydrogen bonds. This causes a transmittance at 1450 whereas $\ce{C-H}$ bonds have a transmittance at 3000.

→ An oxygen atom has a higher molecular mass than hydrogen atoms and this also leads to different frequencies of infrared radiation being absorbed in the one molecule.

(Students could have also discussed the strength of bonds, bond length or molecular vibrations)

Show Worked Solution

→ As infrared radiation is passed through the molecules, the different bonds within the molecule vibrate at specific wavelengths leading to the absorption of the infrared radiation.

→ In this way, different frequencies of infrared radiation can be absorbed by a molecule as bonds differ in electronegativity, dipole strengths and in the masses of atoms at the end of bonds.

→ An oxygen atom has a higher molecular mass than hydrogen atoms and this also leads to different frequencies of infrared radiation being absorbed in the one molecule.

(Students could have also discussed the strength of bonds, bond length or molecular vibrations)

♦♦ Mean mark 30%.
COMMENT: A deep understanding of the principles behind analytical techniques required here.

CHEMISTRY, M8 2013 VCE 2

The strength of the eggshell of birds is determined by the calcium carbonate, $\ce{CaCO3}$, content of the eggshell.

The percentage of calcium carbonate in the eggshell can be determined by gravimetric analysis.

0.412 g of clean, dry eggshell was completely dissolved in a minimum volume of dilute hydrochloric acid.

$\ce{CaCO3(s) + 2H+(aq)\rightarrow Ca^2+(aq) + CO2(g) + H2O(l)}$

An excess of a basic solution of ammonium oxalate, $\ce{(NH4)2C2O4}$, was then added to form crystals of calcium oxalate monohydrate, $\ce{CaC2O4.H2O}$.

The suspension was filtered and the crystals were then dried to constant mass.

0.523 g of $\ce{CaC2O4.H2O}$ was collected.

Write a balanced equation for the formation of the calcium oxalate monohydrate precipitate. (1 mark)

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Determine the percentage, by mass, of calcium carbonate in the eggshell. (3 marks)

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a. $\ce{(NH4)2C2O4(aq) + Ca^2+(aq) + H2O(l) \rightarrow CaC2O4.H2O(s) + 2NH4^+(aq)}$

b. $86.9$%

Show Worked Solution

a. $\ce{(NH4)2C2O4(aq) + Ca^2+(aq) + H2O(l) \rightarrow CaC2O4.H2O(s) + 2NH4^+(aq)}$

♦♦♦ Mean mark (a) 30%.

b. $\ce{M(CaC2O4.H2O)= 40.08 + 2(12.01) + 4(16) + 2(1.008) + 16 = 146.116\ \text{g mol}^{-1}}$

$\ce{n(CaC2O4.H2O)=\dfrac{0.523}{146.116}=0.003579\ \text{mol}}$

$\ce{n(CaC2O4.H2O) = n(Ca^2+)= n(CaCO3) = 0.003579\ \text{mol}}$

$\ce{m(CaCO3)=0.003579 \times (40.08 +12.01 + 3(16))= 0.358\ \text{g}}$

$\text{% Mass}\ =\dfrac{0.358}{0.412} \times 100 = 86.9\%$

CHEMISTRY, M7 2018 VCE 1a

Organic compounds are numerous and diverse due to the nature of the carbon atom. There are international conventions for the naming and representation of organic compounds.

Draw the structural formula of 2-methyl-propan-2-ol. (1 mark)

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Give the molecular formula of but-2-yne. (1 mark)
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Give the IUPAC name of the compound that has the structural formula shown above. (1 mark)

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ii. $\ce{C4H6}$

→ Structural or semi-structural formulas are not appropriate.

iii. → Longest carbon chain is 6, hence hexane

→ Number carbons to give functional group lowest possible numbers: 2,3-dibromo, 4-methyl

→ Compound name: 2,3-dibromo-4-methylhexane

Show Worked Solution

ii. $\ce{C4H6}$

→ Structural or semi-structural formulas are not appropriate.

iii. → Longest carbon chain is 6, hence hexane

→ Number carbons to give functional group lowest possible numbers: 2,3-dibromo, 4-methyl

→ Compound name: 2,3-dibromo-4-methylhexane

♦♦♦ Mean mark (c) 27%.

CHEMISTRY, M8 2016 VCE 6

Brass is an alloy of copper and zinc.

To determine the percentage of copper in a particular sample of brass, an analyst prepared a number of standard solutions of copper$\text{(II)}$ ions and measured their absorbance using an atomic absorption spectrometer (AAS).

The calibration curve obtained is shown below.

A 0.198 g sample of the brass was dissolved in acid and the solution was made up to 100.00 mL in a volumetric flask. The absorbance of this test solution was found to be 0.13
Calculate the percentage by mass of copper in the brass sample. (3 marks)

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If the analyst had made up the solution of the brass sample to 20.00 mL instead of 100.00 mL, would the result of the analysis have been equally reliable? Why? (2 marks)

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Name another analytical technique that could be used to verify the result from part a. (1 mark)

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a. 55.6%

b. → No, this would increase the concentration of the copper solution by a factor of 5.

→ The concentration of the solution and absorbance would be too high and outside the range of the calibration curve (it can’t be assumed that the calibration curve remains linear beyond the range of the known data).

c. Answers could include:

→ UV-vis spectroscopy, colorimetry, volumetric analysis, gravimetric analysis.

Show Worked Solution

a. Absorbance of 0.13 → $\ce{Cu^2+}$ concentration of 1.1 gL$^{-1}$ (see graph)

$\text{In 100 mL:}$

$\ce{m(Cu^2+)}=1.1 \times 0.1=0.11\ \text{g}$

$\Rightarrow \ce{\% Cu^2+}=\dfrac{0.11}{0.198} \times 100=55.6\%$

b. → No, this would increase the concentration of the copper solution by a factor of 5.

→ The concentration of the solution and absorbance would be too high and outside the range of the calibration curve.

→ It can’t be assumed that the calibration curve remains linear beyond the range of the known data.

♦♦♦ Mean mark (b) 25%.

c. Answers could include:

→ UV-vis spectroscopy, colorimetry, volumetric analysis, gravimetric analysis.

PHYSICS, M8 2019 VCE 17

Students are comparing the diffraction patterns produced by electrons and X-rays, in which the same spacing of bands is observed in the patterns, as shown schematically in Figure 18. Note that both patterns shown are to the same scale.

The electron diffraction pattern is produced by 3.0 × 10$^3$ eV electrons.

Explain why electrons can produce the same spacing of bands in a diffraction pattern as X-rays. (3 marks)

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Calculate the frequency of X-rays that would produce the same spacing of bands in a diffraction pattern as for the electrons. Show your working. (4 marks)

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a. → Electrons with momentum exhibit wave-like properties.

→ In this way, moving electrons produce a de Broglie wavelength.

→ As diffraction is a wave phenomenon and is dependent on wavelengths, if the de Broglie wavelength of an electron matches the wavelength of an X-ray then spacing of the bands will be the same.

b. $1.34 \times 10^{19}\ \text{Hz}$

Show Worked Solution

a. → Electrons with momentum exhibit wave-like properties.

→ In this way, moving electrons produce a de Broglie wavelength.

→ As diffraction is a wave phenomenon and is dependent on wavelengths, if the de Broglie wavelength of an electron matches the wavelength of an X-ray then spacing of the bands will be the same.

♦ Mean mark (a) 49%.

b. Find velocity of the electrons using $E=\dfrac{1}{2}mv^2 :$

$3.0 \times 10^3 \times 1.602 \times 10^{-19}=\dfrac{1}{2} \times 9.109 \times 10^{-31} \times v^2$

$v^2$	$=\dfrac{4.806 \times 10^{-16}}{4.5545 \times 10^{-31}}$
$v$	$=\sqrt{1.055 \times 10^{15}}$
	$=3.25 \times 10^7\ \text{ms}^{-1}$

The de Broglie wavelength of the electron is:

$\lambda$	$=\dfrac{h}{mv}$
	$=\dfrac{6.626 \times 10^{-34}}{3.25 \times 10^7 \times 9.109 \times 10^{-31}}$
	$=2.24 \times 10^{-11}\ \text{m}$

Frequency of the X-ray:

$f=\dfrac{c}{\lambda}=\dfrac{3 \times 10^8}{2.24 \times 10^{-11}}=1.34 \times 10^{19}\ \text{Hz}$

♦♦♦ Mean mark (b) 27%.
COMMENT: Multi-step solutions require clear and logical working.

PHYSICS, M5 2019 VCE 8

A 250 g toy car performs a loop in the apparatus shown in Figure 8.

The car starts from rest at point $\text{A}$ and travels along the track without any air resistance or retarding frictional forces. The radius of the car's path in the loop is 0.20 m. When the car reaches point $\text{B}$ it is travelling at a speed of 3.0 m s$^{-1}$.

Calculate the value of $h$. Show your working. (3 marks)

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Calculate the magnitude of the normal reaction force on the car by the track when it is at point $\text{B}$. Show your working. (3 marks)

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Explain why the car does not fall from the track at point $\text{B}$, when it is upside down. (2 marks)

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a. $0.86\ \text{m}$

b. $8.8\ \text{N}$

c. Method 1

→ During the car’s circular motion around the loop, the magnitude of the centripetal force exceeds the magnitude of the weight force.

→ This produces a normal reaction force acting downwards which enables the car to travel around the loop at 3 ms$^{-1}$.

Method 2

→ The minimum speed required for the car so that it stays on the track will be when the centripetal force is equal to the weight force.

$\dfrac{mv^2}{r}$	$=mg$
$v$	$=\sqrt{gr}$
	$=\sqrt{9.8 \times 0.2}$
	$=1.4\ \text{ms}^{-1}$

→ As 3 ms$^{-1}$ is greater than the minimum speed required, a normal force downwards will be present at point $\text{B}$.

→ Therefore the car will not fall from the track at point $\text{B}$.

Show Worked Solution

a. Using the law of the conservation of energy:

$mgh_A$	$=mgh_B +\dfrac{1}{2}mv^2_b$
$0.25 \times 9.8 \times h_A$	$=0.25 \times 9.8 \times 0.4 + \dfrac{1}{2} \times 0.25 \times 3^2$
$2.45h_A$	$=2.105$
$h_A$	$=\dfrac{2.105}{2.45}$
	$=0.86\ \text{m}$

b.	$N + mg$	$=\dfrac{mv^2}{r}$
	$N$	$=\dfrac{mv^2}{r}-mg$
		$=\dfrac{0.25 \times 3^2}{0.2}- 0.25 \times 9.8$
		$=8.8\ \text{N}$

♦ Mean mark (b) 53%.

c. Method 1

→ During the car’s circular motion around the loop, the magnitude of the centripetal force exceeds the magnitude of the weight force.

→ This produces a normal reaction force acting downwards which enables the car to travel around the loop at 3 ms$^{-1}$.

Method 2

→ The minimum speed required for the car so that it stays on the track will be when the centripetal force is equal to the weight force.

$\dfrac{mv^2}{r}$	$=mg$
$v$	$=\sqrt{gr}$
	$=\sqrt{9.8 \times 0.2}$
	$=1.4\ \text{ms}^{-1}$

→ As 3 ms$^{-1}$ is greater than the minimum speed required, a normal force downwards will be present at point $\text{B}$.

→ Therefore the car will not fall from the track at point $\text{B}$.

♦♦♦ Mean mark (c) 23%.
COMMENT: Normal force is poorly understood here..

PHYSICS, M6 2019 VCE 6

A home owner on a large property creates a backyard entertainment area. The entertainment area has a low-voltage lighting system. To operate correctly, the lighting system requires a voltage of 12 V. The lighting system has a resistance of 12 $\Omega$.

Calculate the power drawn by the lighting system. (1 mark)

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To operate the lighting system, the home owner installs an ideal transformer at the house to reduce the voltage from 240 V to 12 V. The home owner then runs a 200 m long heavy-duty outdoor extension lead, which has a total resistance of 3 $ \Omega$, from the transformer to the entertainment area.

The lights are a little dimmer than expected in the entertainment area.
Give one possible reason for this and support your answer with calculations. (4 marks)

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Using the same equipment, what changes could the home owner make to improve the brightness of the lights? Explain your answer. (2 marks)

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a. $12\ \text{W}$

b. Potential cause of dimmer lights:

→ Power loss in the extension lead.

→ The total resistance ($R_T$) in the circuit $=12 + 3=15\ \Omega$

→ The new current in the circuit $=\dfrac{V}{R_T}=\dfrac{12}{15}=0.8\ \text{A}$

→ The power loss across the extension lead, $P_{\text{loss}}=I^2R_l=0.8^2 \times 3=1.92\ \text{W}$

→ Therefore, power supplied to lights $=12.0-1.92=10.08\ \text{W}$.

c. Changes using the same equipment:

→ The homeowner needs to move the transformer from inside the home (start of the lead) to the outside entertainment area (end of the lead).

→ This would mean a higher voltage/lower current is running through the lead which would decrease the power loss through the lead and improve the brightness of the lights.

Show Worked Solution

a. $I=\dfrac{V}{R}=\dfrac{12}{12}=1\ \text{A}$

$P=IV=1 \times 12 = 12\ \text{W}$

b. Potential cause of dimmer lights:

→ Power loss in the extension lead.

→ The total resistance ($R_T$) in the circuit $=12 + 3=15\ \Omega$

→ The new current in the circuit $=\dfrac{V}{R_T}=\dfrac{12}{15}=0.8\ \text{A}$

→ The power loss across the extension lead, $P_{\text{loss}}=I^2R_l=0.8^2 \times 3=1.92\ \text{W}$

→ Therefore, power supplied to lights $=12.0-1.92=10.08\ \text{W}$.

♦♦♦ Mean mark (b) 23%.
COMMENT: Many students used 1 Amp in their calculations instead of finding the total current in the system.

c. Changes using the same equipment:

→ The homeowner needs to move the transformer from inside the home (start of the lead) to the outside entertainment area (end of the lead).

→ This would mean a higher voltage/lower current is running through the lead which would decrease the power loss through the lead and improve the brightness of the lights.

♦♦♦ Mean mark (c) 22%.
COMMENT: Students incorrectly added or changed the equipment used in the question.

PHYSICS, M6 2019 VCE 1

A particle of mass $m$ and charge $q$ travelling at velocity $v$ enters a uniform magnetic field $\text{B}$, as shown in Figure 1.

Is the charge $q$ positive or negative? Give a reason for your answer. (1 mark)

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Explain why the path of the particle is an arc of a circle while the particle is in the magnetic field. (2 marks)

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a. Using the right-hand rule:

→ Thumb is in direction of velocity (right) and fingers are in direction of B-field (Into the page).

→ As the force on the charge is down the page (back of the hand), the charge must be negative.

b. For circular motion to occur:

→ The force must be at right-angles to the velocity of the particle.

→ The force must be of constant magnitude $(F=qvB)$.

→ In the given example, the force on the charged particle due to the magnetic field is perpendicular to its velocity and the magnitude of the force remains constant.

→ The particle will therefore follow the arc of a circle while in the magnetic field.

Show Worked Solution

a. Using the right-hand rule:

→ Thumb is in direction of velocity (right) and fingers are in direction of B-field (Into the page).

→ As the force on the charge is down the page (back of the hand), the charge must be negative.

♦ Mean mark (a) 44%.

b. For circular motion to occur:

→ The force must be at right-angles to the velocity of the particle.

→ The force must be of constant magnitude $(F=qvB)$.

→ In the given example, the force on the charged particle due to the magnetic field is perpendicular to its velocity and the magnitude of the force remains constant.

→ The particle will therefore follow the arc of a circle while in the magnetic field.

♦♦♦ Mean mark (b) 27%.

PHYSICS, M7 2020 VCE 11

An astronaut has left Earth and is travelling on a spaceship at 0.800$c$ directly towards the star known as Sirius, which is located 8.61 light-years away from Earth, as measured by observers on Earth.

How long will the trip take according to a clock that the astronaut is carrying on his spaceship? Show your working. (2 marks)

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Is the trip time measured by the astronaut in part a. a proper time? Explain your reasoning. (2 marks)

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a. $6.46\ \text{years}$

b. → The trip time of 6.46 years on the spaceship is a proper time.

→ This is due to the Astronaut’s clock being stationary within the astronaut’s frame of reference.

Show Worked Solution

a. From the earth’s perspective:

$\text{Travel time}\ =\dfrac{8.61}{0.8}=10.76\ \text{years}$.

From the astronaut’s perspective:

→ The Earth’s time is going to be dilated compared to his, so the time the astronauts clock will measure is:

$t=\dfrac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}$

$t_0$	$=t\sqrt{1-\frac{v^2}{c^2}}$
	$=10.76 \times \sqrt{1-\frac{(0.8c)^2}{c^2}}$
	$=10.76 \times \sqrt{1-0.8^2}$
	$=6.46\ \text{years}$

♦♦♦ Mean mark (a) 29%.
COMMENT: Light years is a measure of distance, not time!

b. → The trip time of 6.46 years on the spaceship is a proper time.

→ This is due to the Astronaut’s clock being stationary within the astronaut’s frame of reference.

♦ Mean mark (b) 44%.

PHYSICS, M6 2020 VCE 6

Two Physics students hold a coil of wire in a constant uniform magnetic field, as shown in Figure 5a. The ends of the wire are connected to a sensitive ammeter. The students then change the shape of the coil by pulling each side of the coil in the horizontal direction, as shown in Figure 5b. They notice a current register on the ammeter.

Will the magnetic flux through the coil increase, decrease or stay the same as the students change the shape of the coil? (1 mark)

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Explain, using physics principles, why the ammeter registered a current in the coil and determine the direction of the induced current. (3 marks

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The students then push each side of the coil together, as shown in Figure 6a, so that the coil returns to its original circular shape, as shown in Figure 6b, and then changes to the shape shown in Figure 6c.

Describe the direction of any induced currents in the coil during these changes. Give your reasoning. (2 marks)

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a. Decrease

b. Ammeter registers a current:

→ By Faraday’s law of induction, a change in flux will result in an induced EMF through the coil and induced current.

→ This induced current acts to oppose the original change in flux that created it. As the magnetic flux is decreasing, the induced current will act to strengthen the magnetic field.

→ Hence by using the right-hand grip rule, the current will run clockwise through the coil.

c. → There will be an induced current from 6a to 6b.

→ As the magnetic flux is increasing, the induced current will act to decrease the flux and oppose the magnetic field.

→ By the right-hand grip rule, the current will flow anti-clockwise around the coil.

→ There will also be an induced current from 6b to 6c.

→ The magnetic flux this time is decreasing, so the induced current will act to increase the flux and strengthen the magnetic field.

→ Using the right-hand grip rule, the current through the coil will flow clockwise.

Show Worked Solution

a. Magnetic flux will decrease.

→ The magnetic flux is proportional to how many field lines are passing through a given area.

→ As the number of lines passing through the coil of wire decreases, the magnetic flux will also decrease.

b. Ammeter registers a current:

→ By Faraday’s law of induction, a change in flux will result in an induced EMF through the coil and induced current.

→ This induced current acts to oppose the original change in flux that created it. As the magnetic flux is decreasing, the induced current will act to strengthen the magnetic field.

→ Hence by using the right-hand grip rule, the current will run clockwise through the coil.

♦ Mean mark (b) 42%.

c. → There will be an induced current from 6a to 6b.

→ As the magnetic flux is increasing, the induced current will act to decrease the flux and oppose the magnetic field.

→ By the right-hand grip rule, the current will flow anti-clockwise around the coil.

→ There will also be an induced current from 6b to 6c.

→ The magnetic flux this time is decreasing, so the induced current will act to increase the flux and strengthen the magnetic field.

→ Using the right-hand grip rule, the current through the coil will flow clockwise.

♦♦♦ Mean mark (c) 28%.
COMMENT: Lenz’s law was poorly understood in this question.

PHYSICS, M5 2020 VCE 4*

The Ionospheric Connection Explorer (ICON) space weather satellite, constructed to study Earth's ionosphere, was launched in October 2019. ICON will study the link between space weather and Earth's weather at its orbital altitude of 600 km above Earth's surface. Assume that ICON's orbit is a circular orbit.

Calculate the orbital radius of the ICON satellite. (1 mark)

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Calculate the orbital period of the ICON satellite correct to three significant figures. Show your working. (4 marks)

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Explain how the ICON satellite maintains a stable circular orbit without the use of propulsion engines. (2 marks)

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a. $r_{orbit}= 6.971 \times 10^6\ \text{m}$

b. $5780\ \text{s}$

c. → The icon satellite is only subject to the gravitational force of the Earth.

→ This force is a centripetal force of constant magnitude and hence the satellite maintains a stable circular orbit.

Show Worked Solution

a. Radius of orbit is equal to the altitude plus the radius of the Earth.

$r_{orbit}=600\ 000\ \text{m} + 6.371 \times 10^6\ \text{m} = 6.971 \times 10^6\ \text{m}$

b.	$\dfrac{r^3}{T^2}$	$=\dfrac{GM}{4\pi^2}$
	$T$	$=\sqrt{\dfrac{4\pi^2r^3}{GM}}$
		$=\sqrt{\dfrac{4\pi^2 \times (6.971 \times 10^6)^3}{6.67 \times 10^{-11} \times 6 \times 10^{24}}}$
		$=5780\ \text{s}\ \ \text{(3 sig.fig.)}$

c. → The icon satellite is only subject to the gravitational force of the Earth.

→ This force is a centripetal force of constant magnitude and hence the satellite maintains a stable circular orbit.

♦♦♦ Mean mark (c) 23%.

PHYSICS, M6 2020 VCE 3

Electron microscopes use a high-precision electron velocity selector consisting of an electric field, $E$, perpendicular to a magnetic field, $B$.

Electrons travelling at the required velocity, $v_0$, exit the aperture at point $\text{Y}$, while electrons travelling slower or faster than the required velocity, $v_0$, hit the aperture plate, as shown in Figure 2.

Show that the velocity of an electron that travels straight through the aperture to point $\text{Y}$ is given by $ v_{0} $ = $ \dfrac{E}{B}$. (1 mark)

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Calculate the magnitude of the velocity, $v_0$, of an electron that travels straight through the aperture to point $\text{Y}$ if $E$ = 500 kV m$^{-1}$ and $B$ = 0.25 T. Show your working. (2 marks)

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i. At which of the points – $\text{X, Y}$, or $\text{Z}$ – in Figure 2 could electrons travelling faster than $v_0$ arrive? (1 mark)

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ii. Explain your answer to part c.i. (2 marks)

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a. Find $v_0$ where forces due to magnetic and electric field are balanced

$F_E$	$=F_B$
$qE$	$=qv_0B$
$v_0$	$=\dfrac{E}{B}$

b. $v_0=2 \times 10^6\ \text{ms}^{-1}$

c.i. Point $\text{Z.}$

c.ii. When electrons travel faster than $v_0$:

→ The force due to the magnetic field will increase but the force due to the electric field will remain unchanged.

→ Therefore, there will be a net force acting on the electron due to the magnetic force being greater than the electric force.

→ Using the right-hand rule, the force on the electron due to the magnetic field is down the page, hence the electron will arrive at point $\text{Z.}$

Show Worked Solution

a. Find $v_0$ where forces due to magnetic and electric field are balanced

$F_E$	$=F_B$
$qE$	$=qv_0B$
$v_0$	$=\dfrac{E}{B}$

♦ Mean mark (a) 46%.

b. $v_0=\dfrac{E}{B}=\dfrac{500\ 000}{0.25}=2 \times 10^6\ \text{ms}^{-1}$

c.i. Point $\text{Z.}$

c.ii. When electrons travel faster than $v_0$:

→ The force due to the magnetic field will increase but the force due to the electric field will remain unchanged.

→ Therefore, there will be a net force acting on the electron due to the magnetic force being greater than the electric force.

→ Using the right-hand rule, the force on the electron due to the magnetic field is down the page, hence the electron will arrive at point $\text{Z.}$

♦ Mean mark (c.i.) 40%.
♦♦♦ Mean mark (c.ii.) 15%.
COMMENT: Students needed to include what would happen to the electric force.

CHEMISTRY, M3 EQ-Bank 12

A student stirs 2.80 g of silver $\text{(I)}$ nitrate powder into 250.0 mL of 1.00 mol L$^{-1}$ sodium hydroxide solution until it is fully dissolved. A reaction occurs and a precipitate appears.

Write a balanced chemical equation for the reaction. (1 mark)

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Calculate the theoretical mass of precipitate that will be formed. (3 marks)

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The student weighed a piece of filter paper, filtered out the precipitate and dried it thoroughly in an incubator. The final precipitate mass was higher than predicted in (b).

Identify one scientific reason why the precipitate mass was too high and suggest an improvement to the experimental method which would eliminate this error. (2 marks)

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a. $\ce{AgNO3(aq) + NaOH(aq) -> AgOH(s) + NaNO3(aq)}$

b. $2.06 \text{ g}$

c. Possible reasons for the higher mass:

→ Presence of excess solution in the filter paper or incomplete drying.

→ Presence of soluble ions on the filter paper which would crystalise when dried and increase the mass of the precipitate.

Experimental adjustment to eliminate error:

→ Ensure the precipitate is thoroughly rinsed with distilled water to remove soluble impurities.

→ Dry precipitate completely in the incubator before weighing.

Show Worked Solution

a. $\ce{AgNO3(aq) + NaOH(aq) -> AgOH(s) + NaNO3(aq)}$

b. $\ce{MM(AgNO3) = 107.9 + 14.01 + 16.00 \times 3 = 169.91}$

$\ce{n(AgNO3) = \dfrac{\text{m}}{\text{MM}} = \dfrac{2.80}{169.91} = 0.01648 \text{ mol}}$

$\ce{n(NaOH) = c \times V = 1.00 \times 0.250 = 0.250 \text{ mol}}$

$\Rightarrow \ce{AgNO3}\ \text{is the limiting reagent}$

$\ce{n(AgOH) = n(AgNO3) = 0.01648 \text{ mol}}$

$\ce{m(AgOH) = n \times MM = 0.01648 \times (107.9 + 16.00 + 1.008) = 2.06 \text{ g}}$

c. Possible reasons for the higher mass:

→ Presence of excess solution in the filter paper or incomplete drying.

→ Presence of soluble ions on the filter paper which would crystalise when dried and increase the mass of the precipitate.

Experimental adjustment to eliminate error:

→ Ensure the precipitate is thoroughly rinsed with distilled water to remove soluble impurities.

→ Dry precipitate completely in the incubator before weighing.

\(S_{n+1}\)	\(=TS_n+A\)
\(TS_n\)	\(=S_{n+1}-A\)
\(S_n\)	\(=T^{-1}\times \left(S_{n+1}-A\right)\)

	\(\dfrac{g-k v_0}{g+k v_{\small{A}}}\)	\(=e^{kt}\)
	\(g-kv_0\)	\(=e^{kt} \cdot g+e^{kt} \cdot kv_{\small{A}}\)
	\(kv_{\small{A}}\)	\(=e^{-kt}\left(g-kv_0\right)-g\)
	\(v_{\small{A}}\)	\(=\dfrac{1}{k}\left[e^{-kt}\left(g-kv_0\right)-g\right]\)

	\(x\)	\(=\displaystyle \frac{1}{k} \int e^{-kt} \cdot g-e^{-kt} \cdot v_0-g \, dt\)
		\(=\dfrac{1}{k}\left[-\dfrac{g}{k}e^{-kt}+v_0 e^{-kt}-gt\right]+c\)
		\(=-\left(\dfrac{g}{k^2}-\dfrac{v_0}{k}\right) e^{-kt}-gt+c\)

	\(v_{\small{B}}\)	\(=\dfrac{1}{k}\left[e^{-kt}\left(g+kv_0\right)-g\right]\)
	\(x_{\small{B}}\)	\(=-\left(\dfrac{g}{k^2}+\dfrac{v_0}{k}\right) e^{-kt}-gt+c\)

	\(\dfrac{2 v_0}{k} \cdot e^{-k t}\)	\(=\dfrac{2 v_0}{k}-d\)
	\(e^{-kt}\)	\(=\left(\dfrac{2 v_0-d k}{k}\right) \cdot \dfrac{k}{v_0}\)
	\(-kt\)	\(=\ln \left(\dfrac{2 v_0-d k}{v_0}\right)\)
	\(t\)	\(=-\dfrac{1}{k} \ln \left(\dfrac{2v_0-dk}{v_0}\right)\)

	\((\gamma+\bar{\gamma})_1\)	\(=2 \cos \left(\dfrac{2 \pi}{9}\right)\)
	\((\gamma+\bar{\gamma})_2\)	\(=2 \cos \left(\dfrac{8 \pi}{9}\right)\)
	\((\gamma+\bar{\gamma})_3\)	\(=2 \cos \left(\dfrac{4\pi}{9}\right)\)

	\((\gamma+\bar{\gamma})_1\)	\(=2 \cos \left(\dfrac{2 \pi}{9}\right)\)
	\((\gamma+\bar{\gamma})_2\)	\(=2 \cos \left(\dfrac{8 \pi}{9}\right)\)
	\((\gamma+\bar{\gamma})_3\)	\(=2 \cos \left(\dfrac{4\pi}{9}\right)\)

	\(\overrightarrow{O R}\)	\(=\overrightarrow{O P}+k \overrightarrow{P Q}\)
		\(=\overrightarrow{O P}+k(\overrightarrow{O Q}-\overrightarrow{O P})\)
		\(=(1-k) \overrightarrow{O P}+k \overrightarrow{O Q}\)
		\(=\dfrac{3}{5}(1-k)\underset{\sim}{a}+3 k \underset{\sim}{b}\)

	\(\dfrac{3}{5}(1-k)\)	\(=1-3 k\)
	\(3-3 k\)	\(=5-15 k\)
	\(k\)	\(=\dfrac{1}{6}\)