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PHYSICS, M6 2021 VCE 6

Figure 6 shows a simple AC generator. A mechanical energy source rotates the loop smoothly at 50 revolutions per second and the loop generates a voltage of 6 V. The magnetic field, \(B\), is constant and uniform. The direction of rotation is as shown in Figure 6.
 

  1. Sketch the output EMF between \(\text{P}\) and \(\text{Q}\) versus time, \(t\), on the grid below, starting with the loop in the position shown in Figure 6. Show at least two complete revolutions, and include the voltage on the vertical axis and a time scale on the horizontal axis.   (3 marks)
     

  1. Describe the function of the slip rings shown in Figure 6.   (1 mark)

  1.  i. How could the AC generator shown in Figure 6 be changed to a DC generator?   (1 mark)

 ii.  Sketch the output EMF versus time, \(t\), for this DC generator for at least two complete revolutions on the grid below. Include the voltage on the vertical axis and a time scale on the horizontal axis. (2 marks)
 

Show Answers Only

a.   

         

b.    The function of the slip rings:

→ Produce an AC voltage output by maintaining a constant connection between the loop and the external circuit.
 

c.i.   Changing the slip rings to a split ring commutator.

c.ii. 

         

Show Worked Solution

a.    \(\text{Period}\ (T)=\dfrac{1}{f}=\dfrac{1}{50}=0.02\ \text{s}\).

b.    The function of the slip rings:

→ Produce an AC voltage output by maintaining a constant connection between the loop and the external circuit.

♦ Mean mark (b) 54%.

c.i.   Changing the slip rings to a split ring commutator.
 

c.ii.  The output of a DC generator is positive only.

   

♦ Mean mark (c)(i) 55%.

Calculus, SPEC2 2022 VCAA 9 MC

Euler's method is used to find an approximate solution to the differential equation  `\frac{d y}{d x}=2 x^2`.

Given that  `x_0=1, y_0=2`  and  `y_2=2.976`, the value of the step size `h` is

  1. 0.1
  2. 0.2
  3. 0.3
  4. 0.4
  5. 0.5
Show Answers Only

`B`

Show Worked Solution

`y_{n+1}=y_{n}+y_{n}^{\prime}*h`

`y_1=y_0+y_0^{′}*h = 2+2 xx 1^2 * h=2+2h`

`y_2=y_1+y_1^{′}*h = (2+2h)+2 xx (1+h)^2*h = 2h(1+h)^2+2(1+h)`

`text{Solve:}\ 2h(1+h)^2+2(1+h)=2.976,\ \text{for}\ h:`

`h=0.2`

`=>B`

Calculus, SPEC2 2022 VCAA 8 MC

The direction field shown above best represents the differential equation

  1. `\frac{d y}{d x}=\frac{2 x}{y}`
  2. `\frac{d y}{d x}=-\frac{x}{2 y}`
  3. `\frac{d y}{d x}=-\frac{2 x}{y}`
  4. `\frac{d y}{d x}=\frac{y^2}{2}+x^2`
  5. `\frac{d y}{d x}=\frac{x^2}{2}+y^2`
Show Answers Only

`C`

Show Worked Solution

`text{Slope lines are vertical when}\ y=0\ \text{(eliminate D and E)}`

`text{Slope lines are negative in 1st/3rd quadrants (eliminate A)}`

`text{Consider slope at point (1, 1) ⇒ gradient close to –2.}`

`=>C`

Calculus, EXT2 C1 2022 SPEC2 7 MC

Using the substitution  `u=1+e^x, \int_0^{\log _e 2} \frac{1}{1+e^x}dx`  can be expressed as

  1. `\int_0^{\log _e 2}\left(\frac{1}{u-1}-\frac{1}{u}\right) du`
  2. `\int_2^3\left(\frac{1}{u}-\frac{1}{u-1}\right) du`
  3. `\int_1^3\left(\frac{1}{u}-\frac{1}{u-1}\right) du`
  4. `\int_2^3\left(\frac{1}{u-1}-\frac{1}{u}\right) du`
Show Answers Only

`D`

Show Worked Solution

`u=1+e^x\ \ =>\ \ \frac{du}{dx}=e^x\ \ =>\ \ dx=\frac{du}{e^x}=\frac{du}{u-1}`

`text{When}\ \ x=0,\ \ u=2`

`text{When}\ \ x=\log_e 2,\ \ u=1+e^{\log_e 2} = 3`

`\int_0^{\log _e 2} \frac{1}{1+e^x}dx = \int_2^3\left(\frac{1}{u}*\frac{1}{u-1}\right) du= \int_2^{1+e^2}\left(\frac{1}{u-1}-\frac{1}{u}\right) du`

`=>D`

Calculus, SPEC2 2022 VCAA 7 MC

Using the substitution  `u=1+e^x, \int_0^{\log _e 2} \frac{1}{1+e^x}dx`  can be expressed as

  1. `\int_0^{\log _e 2}\left(\frac{1}{u-1}-\frac{1}{u}\right) du`
  2. `\int_2^3\left(\frac{1}{u}-\frac{1}{u-1}\right) du`
  3. `\int_1^3\left(\frac{1}{u}-\frac{1}{u-1}\right) du`
  4. `\int_2^3\left(\frac{1}{u-1}-\frac{1}{u}\right) du`
  5. `\int_2^{1+e^2}\left(\frac{1}{u-1}-\frac{1}{u}\right) du`
Show Answers Only

`D`

Show Worked Solution

`u=1+e^x\ \ =>\ \ \frac{du}{dx}=e^x\ \ =>\ \ dx=\frac{du}{e^x}=\frac{du}{u-1}`

`text{When}\ \ x=0,\ \ u=2`

`text{When}\ \ x=\log_e 2,\ \ u=1+e^{\log_e 2} = 3`

`\int_0^{\log _e 2} \frac{1}{1+e^x}dx = \int_2^3\left(\frac{1}{u}*\frac{1}{u-1}\right) du= \int_2^{1+e^2}\left(\frac{1}{u-1}-\frac{1}{u}\right) du`

`=>D`

Complex Numbers, SPEC2 2022 VCAA 6 MC

Given  `z=x+yi`, where `x, y \in R` and `z \in C`, an equation that has a graph that has two points of intersection with the graph given by  `|z-5|=2`  is

  1. `\text{Arg}(z-3)=\frac{\pi}{2}`
  2. `|z-1|=2`
  3. `\text{Im}(z)=2`
  4. `\text{Re}(z)+\text{Im}(z)=2`
  5. `|z-5-5 i|=4`
Show Answers Only

`E`

Show Worked Solution

`text{Using CAS, sketch all graphs together with the circle}\ |z-5|=2`

`|z-5-5 i|=4\ \ text{has 2 points of intersection.}`

`=>E`

Calculus, MET2 2023 VCAA 3

Consider the function \(g:R \to R, g(x)=2^x+5\).

  1. State the value of \(\lim\limits_{x\to -\infty} g(x)\).   (1 mark)
  2. The derivative, \(g^{'}(x)\), can be expressed in the form \(g^{'}(x)=k\times 2^x\).
  3. Find the real number \(k\).   (1 mark)
  4.  i. Let \(a\) be a real number. Find, in terms of \(a\), the equation of the tangent to \(g\) at the point \(\big(a, g(a)\big)\).   (1 mark)

    ii. Hence, or otherwise, find the equation of the tangent to \(g\) that passes through the origin, correct to three decimal places.   (2 marks)

  1.  

Let \(h:R\to R, h(x)=2^x-x^2\).

  1. Find the coordinates of the point of inflection for \(h\), correct to two decimal places.   (1 mark)
  2. Find the largest interval of \(x\) values for which \(h\) is strictly decreasing.
  3. Give your answer correct to two decimal places.   (1 mark)
  4. Apply Newton's method, with an initial estimate of \(x_0=0\), to find an approximate \(x\)-intercept of \(h\).
  5. Write the estimates \(x_1, x_2,\) and \(x_3\) in the table below, correct to three decimal places.   (2 marks)
      

    \begin{array} {|c|c|}
    \hline
    \rule{0pt}{2.5ex} \qquad x_0\qquad \ \rule[-1ex]{0pt}{0pt} & \qquad \qquad 0 \qquad\qquad \\
    \hline
    \rule{0pt}{2.5ex} x_1 \rule[-1ex]{0pt}{0pt} &  \\
    \hline
    \rule{0pt}{2.5ex} x_2 \rule[-1ex]{0pt}{0pt} &  \\
    \hline
    \rule{0pt}{2.5ex} x_3 \rule[-1ex]{0pt}{0pt} &  \\
    \hline
    \end{array}
  6. For the function \(h\), explain why a solution to the equation \(\log_e(2)\times (2^x)-2x=0\) should not be used as an initial estimate \(x_0\) in Newton's method.   (1 mark)
  7. There is a positive real number \(n\) for which the function \(f(x)=n^x-x^n\) has a local minimum on the \(x\)-axis.
  8. Find this value of \(n\).   (2 marks)
Show Answers Only

a.    \(5\)

b.    \(\log_{e}{2}\ \ \text{or}\ \ \ln\ 2\)

ci.  \(y=2^a\ \log_{e}{(2)x}-(a\ \log_{e}{(2)}-1)\times2^a+5\)

\(\text{or}\ \ y=2^a\ \log_{e}{(2)x}-a\ 2^a\ \log_{e}{(2)}+2^a+5\)

cii. \(y=4.255x\)

d. \((2.06 , -0.07)\)

e. \([0.49, 3.21]\)

f. 

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} \qquad x_0\qquad \ \rule[-1ex]{0pt}{0pt} & \qquad \qquad 0 \qquad\qquad \\
\hline
\rule{0pt}{2.5ex} x_1 \rule[-1ex]{0pt}{0pt} &  -1.433 \\
\hline
\rule{0pt}{2.5ex} x_2 \rule[-1ex]{0pt}{0pt} &  -0.897 \\
\hline
\rule{0pt}{2.5ex} x_3 \rule[-1ex]{0pt}{0pt} &  -0.773 \\
\hline
\end{array}

g. \(\text{See worked solution.}\)

h. \(n=e\)

Show Worked Solution

a.    \(\text{As }x\to -\infty,\ \ 2^x\to 0\)

\(\therefore\ 2^x+5\to 5\)
  

b.     \(g(x)\) \(=2^x+5\)
    \(=\Big(e^{\log_{e}{2}}\Big)^x\)
    \(=e\ ^{x\log_{e}{2}}+5\)
  \(g^{\prime}(x)\) \(=\log_{e}{2}\times e\ ^{x\log_{e}{2}}\)
     \(=\log_{e}{2}\times 2^x\)
  \(\therefore\ k\) \(=\log_{e}{2}\ \ \text{or}\ \ \ln\ 2\)
   
ci.  \(\text{Tangent at}\ (a, g(a)):\)

\(y-(2^a+5)\) \(=\log_{e}{2}\times2^a(x-a)\)
\(\therefore\ y\) \(=2^a\ \log_{e}{(2)x}-a\ 2^a\ \log_{e}{(2)}+2^a+5\)

♦♦ Mean mark (c)(i) 50%.

cii.  \(\text{Substitute }(0, 0)\ \text{into equation from c(i) to find}\ a\)

\( y\) \(=2^a\ \log_{e}{(2)x}-a\ 2^a\ \log_{e}{(2)}+2^a+5\)
\(0\) \(=2^a\ \log_{e}{(2)\times 0}-a\ 2^a\ \log_{e}{(2)}+2^a+5\)
\(0\) \(=-a\ 2^a\ \log_{e}{(2)}+2^a+5\)

  
\(\text{Solve for }a\text{ using CAS }\rightarrow\ a\approx 2.61784\dots\)

\(\text{Equation of tangent when }\ a\approx 2.6178\)

\( y\) \(=2^{2.6178..}\ \log_{e}{(2)x}+0\)
\(\therefore\  y\) \(=4.255x\)

♦♦♦ Mean mark (c)(ii) 30%.
MARKER’S COMMENT: Some students did not substitute (0, 0) into the correct equation or did not find the value of \(a\) and used \(a=0\).
d.     \(h(x)\) \(=2^x-x^2\)
  \(h^{\prime}(x)\) \(=\log_{e}{(2)}\cdot 2^x-2x\ \ \text{(Using CAS)}\)
  \(h^{”}(x)\) \(=(\log_{e}{(2)})^2\cdot 2^x-2\ \ \text{(Using CAS)}\)

\(\text{Solving }h^{”}(x)=0\ \text{using CAS }\rightarrow\ x\approx 2.05753\dots\)

\(\text{Substituting into }h(x)\ \rightarrow\ h(2.05753\dots)\approx-0.070703\dots\)

\(\therefore\ \text{Point of inflection at }(2.06 , -0.07)\ \text{ correct to 2 decimal places.}\)

e.    \(\text{From graph (CAS), }h(x)\ \text{is strictly decreasing between the 2 turning points.}\)

\(\therefore\ \text{Largest interval includes endpoints and is given by }\rightarrow\ [0.49, 3.21]\)


♦♦ Mean mark (e) 40%.
MARKER’S COMMENT: Round brackets were often used which were incorrect as endpoints were included. Some responses showed the interval where the function was strictly increasing.

f.    \(\text{Newton’s Method }\Rightarrow\  x_a-\dfrac{h(x_a)}{h'(x_a)}\) 

\(\text{for }a=0, 1, 2, 3\ \text{given an initial estimation for }x_0=0\)

\(h(x)=2x-x^2\ \text{and }h^{\prime}(x)=\ln{2}\times 2^x-2x\)

\begin{array} {|c|l|}
\hline
\rule{0pt}{2.5ex} \qquad x_0\qquad \ \rule[-1ex]{0pt}{0pt} & \qquad \qquad 0 \qquad\qquad \\
\hline
\rule{0pt}{2.5ex} x_1 \rule[-1ex]{0pt}{0pt} &  0-\dfrac{2^0-2\times 0}{\ln2\times 2^0\times 0}=-1.433 \\
\hline
\rule{0pt}{2.5ex} x_2 \rule[-1ex]{0pt}{0pt} &  -1.433-\dfrac{2^{-1.433}-2\times -1.433}{\ln2\times 2^{-1.433}\times -1.433}=-0.897 \\
\hline
\rule{0pt}{2.5ex} x_3 \rule[-1ex]{0pt}{0pt} &  -0.897-\dfrac{2^{-0.897}-2\times -0.897}{\ln2\times 2^{-0.897}\times -0.897}=-0.773 \\
\hline
\end{array}

g.    \(\text{The denominator in Newton’s Method is}\ h^{\prime}(x)=\log_{e}{(2)}\cdot 2^x-2x\)

\(\text{and the calculation will be undefined if }h^{\prime}(x)=0\ \text{as the tangent lines are horizontal}.\)

\(\therefore\ \text{The solution to }h^{\prime}(x)=0\ \text{cannot be used for }x_0.\)

♦♦♦ Mean mark (g) 20%.

h.    \(\text{For a local minimum }f(x)=0\)

\(\rightarrow\ n^x-x^n=0\)

\(\rightarrow\ n^x=x^n\ \ \ (1)\)

\(\text{Also for a local minimum }f^{\prime}(x)=0\)

\(\rightarrow\ \ln(n)\cdot n^x-nx^{n-1}=0\ \ \ (2)\)

\(\text{Substitute (1) into (2)}\)

\(\ln(n)\cdot x^n-nx^{n-1}=0\) 

\(x^n\Big(\ln(n)-\dfrac{n}{x}\Big)=0\)

\(\therefore\ x^n=0\ \text{or }\ \ln(n)=\dfrac{n}{x}\)

\(x=0\ \text{or }x=\dfrac{n}{\ln(n)}\)

\(\therefore\ n=e\)


♦♦♦ Mean mark (h) 10%.
MARKER’S COMMENT: Many students showed that \(f'(x)=0\) but failed to couple it with \(f(x)=0\). Ensure exact values are given where indicated not approximations.

Algebra, SPEC2 2022 VCAA 2 MC

The expression `1-\frac{4\sin^2(x)}{\tan^2(x)+1}` simplifies to

  1. `sin(x) \cos(x)`
  2. `1-2\cos^2(2x)`
  3. `2\sin(2x)`
  4. `2\sin^2(2x)`
  5. `\cos^2(2x)`
Show Answers Only

`E`

Show Worked Solution
`1-\frac{4 \sin ^2 x}{\tan ^2 x+1}` `= 1-\frac{4 \sin ^2 x}{\sec ^2 x}`  
  `= 1-(2 \sin x\ \cos x)^2`  
  `= 1-\sin ^2 (2 x)`  
  `= \cos ^2 (2 x)`  

 
`=>E`

Mechanics, EXT2 M1 2023 SPEC1 8

A body moves in a straight line so that when its displacement from a fixed origin `O` is `x` metres, its acceleration, `a`, is `-4 x \ text{ms}^{-2}`. The body accelerates from rest and its velocity, `v`, is equal to `-2 \ text{ms}^{-1}` as it passes through the origin. The body then comes to rest again.

Find `v` in terms of `x` for this interval.  (4 marks)

Show Answers Only

` – 2sqrt(1-x^2)`

Show Worked Solution

`a = -4x`

`d/dx(1/2v^2)` `= -4x`  
`1/2v^2` `= -2x^2 +c`  

 
`v= -2\ \ \text{when}\ \ x = 0\ \ =>\ \ c = 2`

`v^2` `= -4x^2 + 4`  
`v^2` `= 4(1-x^2)`  

 
`v= -2\ \ text{when}\ \ x=0:`

`:.\ v` `= -sqrt(4(1-x^2))`  
  `= -2sqrt(1-x^2)`  

Calculus, SPEC1 2022 VCAA 8

A body moves in a straight line so that when its displacement from a fixed origin `O` is `x` metres, its acceleration, `a`, is `-4 x \ text{ms}^{-2}`. The body accelerates from rest and its velocity, `v`, is equal to `-2 \ text{ms}^{-1}` as it passes through the origin. The body then comes to rest again.

Find `v` in terms of `x` for this interval.  (4 marks)

Show Answers Only

` – 2sqrt(1-x^2)`

Show Worked Solution

`a = -4x`

`d/dx(1/2v^2)` `= -4x`  
`1/2v^2` `= -2x^2 +c`  

 
`v= -2\ \ \text{when}\ \ x = 0\ \ =>\ \ c = 2`

`v^2` `= -4x^2 + 4`  
`v^2` `= 4(1-x^2)`  

 
`v= -2\ \ text{when}\ \ x=0:`

`:.\ v` `= -sqrt(4(1-x^2))`  
  `= -2sqrt(1-x^2)`  

Vectors, EXT1 V1 2022 SPEC1 6b

`OPQ` is a semicircle of radius `a` with equation  `y=sqrt(a^(2)-(x-a)^(2))`. `P(x,y)` is a point on the semicircle `OPQ`, as shown below.

  1. Express the vectors `vec(OP)` and `vec(QP)` in terms of  `a`, `x`, `y`, `underset~i` and `underset~j`, where `underset~i` is a unit vector in the direction of the positive `x`-axis and `underset~j` is a unit vector in the direction of the positive `y`-axis.   (1 mark)

  2. Hence, using the vector scalar (dot) product, determine whether `vec(OP)` is perpendicular to `vec(QP)`.   (3 marks)

Show Answers Only

i.    `vec(OP)= xunderset~i + sqrt(a^2-(x-a)^2 underset~j`

`vec(QP)= (x-2a) underset~i + sqrt(a^2-(x-a)^2) underset~j`
 

ii.   `vec(OP)* vec(QP)` `=x(x-2a)+a^(2)-(x-a)^(2)`
  `=x^(2)-2ax+a^(2)-x^(2)+2ax-a^(2)`
  `= 0`

 
`:. vec(OP)\ \text{and}\ vec(QP)\ \text{are perpendicular.}`

Show Worked Solution

i.    `vec(OP)= xunderset~i + sqrt(a^2-(x-a)^2 underset~j`

`vec(QP)= (x-2a) underset~i + sqrt(a^2-(x-a)^2) underset~j`
 

ii.   `vec(OP)* vec(QP)` `=x(x-2a)+a^(2)-(x-a)^(2)`
  `=x^(2)-2ax+a^(2)-x^(2)+2ax-a^(2)`
  `= 0`

 
`:. vec(OP)\ \text{and}\ vec(QP)\ \text{are perpendicular.}`

Vectors, SPEC1 2022 VCAA 6b

`OPQ` is a semicircle of radius `a` with equation  `y=sqrt(a^(2)-(x-a)^(2))`. `P(x,y)` is a point on the semicircle `OPQ`, as shown below.

  1. Express the vectors `vec(OP)` and `vec(QP)` in terms of  `a`, `x`, `y`, `underset~i` and `underset~j`, where `underset~i` is a unit vector in the direction of the positive `x`-axis and `underset~j` is a unit vector in the direction of the positive `y`-axis.   (1 mark)

  2. Hence, using the vector scalar (dot) product, determine whether `vec(OP)` is perpendicular to `vec(QP)`.   (3 marks)

Show Answers Only

i.    `vec(OP)= xunderset~i + sqrt(a^2-(x-a)^2 underset~j`

`vec(QP)= (x-2a) underset~i + sqrt(a^2-(x-a)^2) underset~j`
 

ii.   `vec(OP)* vec(QP)` `=x(x-2a)+a^(2)-(x-a)^(2)`
  `=x^(2)-2ax+a^(2)-x^(2)+2ax-a^(2)`
  `= 0`

 
`:. vec(OP)\ \text{and}\ vec(QP)\ \text{are perpendicular.}`

Show Worked Solution

i.    `vec(OP)= xunderset~i + sqrt(a^2-(x-a)^2 underset~j`

`vec(QP)= (x-2a) underset~i + sqrt(a^2-(x-a)^2) underset~j`
 

ii.   `vec(OP)* vec(QP)` `=x(x-2a)+a^(2)-(x-a)^(2)`
  `=x^(2)-2ax+a^(2)-x^(2)+2ax-a^(2)`
  `= 0`

 
`:. vec(OP)\ \text{and}\ vec(QP)\ \text{are perpendicular.}`

Calculus, EXT2 C1 2022 SPEC1 4

Find `int(3x^(2)+4x+12)/(x(x^(2)+4))\ dx`.   (4 marks)

Show Answers Only

`3ln |x|+2tan^(-1)((x)/(2))+c`

Show Worked Solution

`text{Using partial fractions:}`

`(3x^(2)+4x+12)/(x(x^(2)+4))` `-=(A)/(x)+(Bx+C)/(x^(2)+4)`  
`3x^(2)+4x+12` `=A(x^(2)+4)+x(Bx+C)`  
`3x^(2)+4x+12` `=(A+B)x^(2)+Cx+4A`  

 
`4A=12\ \ =>\ \ A=3`

`C=4`

`A+B=3\ \ =>\ \ B=0`

`:.\int(3x^(2)+4x+12)/(x(x^(2)+4))\ dx` `=int(3)/(x)+(4)/(4+x^(2))\ dx`  
  `=3ln |x|+2tan^(-1)((x)/(2))+c`  
Mean mark 55%.

Calculus, SPEC1 2022 VCAA 4

Find `int(3x^(2)+4x+12)/(x(x^(2)+4))\ dx`.   (4 marks)

Show Answers Only

`3ln |x|+2tan^(-1)((x)/(2))+c`

Show Worked Solution

`text{Using partial fractions:}`

`(3x^(2)+4x+12)/(x(x^(2)+4))` `-=(A)/(x)+(Bx+C)/(x^(2)+4)`  
`3x^(2)+4x+12` `=A(x^(2)+4)+x(Bx+C)`  
`3x^(2)+4x+12` `=(A+B)x^(2)+Cx+4A`  

 
`4A=12\ \ =>\ \ A=3`

`C=4`

`A+B=3\ \ =>\ \ B=0`

`:.\int(3x^(2)+4x+12)/(x(x^(2)+4))\ dx` `=int(3)/(x)+(4)/(4+x^(2))\ dx`  
  `=3ln |x|+2tan^(-1)((x)/(2))+c`  
Mean mark 55%.

Calculus, EXT1 C3 2022 SPEC1 2

Solve the differential equation  `(dy)/(dx) = -x sqrt(4-y^2)`  given that  `y(2) = 0`. Give your answer in the form  `y = f(x)`.   (3 marks)

Show Answers Only

`y=2sin(-(1)/(2)x^(2)+2)`

Show Worked Solution
`int(dy)/(sqrt(4-y^(2)))` `=int-x\ dx`  
`sin^(-1)((y)/(2))` `=-(1)/(2)x^(2)+c`  

 
`y(2)=0\ \=> \ c=2`

`(y)/(2)` `=sin(-(1)/(2)x^(2)+2)`  
`y` `=2sin(-(1)/(2)x^(2)+2)`  

Calculus, SPEC1 2022 VCAA 2

Solve the differential equation  `(dy)/(dx) = -x sqrt(4-y^2)`  given that  `y(2) = 0`. Give your answer in the form  `y = f(x)`.   (3 marks)

Show Answers Only

`y=2sin(-(1)/(2)x^(2)+2)`

Show Worked Solution
`int(dy)/(sqrt(4-y^(2)))` `=int-x\ dx`  
`sin^(-1)((y)/(2))` `=-(1)/(2)x^(2)+c`  

 
`y(2)=0\ \=> \ c=2`

`(y)/(2)` `=sin(-(1)/(2)x^(2)+2)`  
`y` `=2sin(-(1)/(2)x^(2)+2)`  

Vectors, SPEC2 2023 VCAA 5

The points with coordinates \(A(1,1,2), B(1,2,3)\) and \(C(3,2,4)\) all lie in a plane \(\Pi\).

  1. Find the vectors \(\overrightarrow{A B}\) and \(\overrightarrow{A C}\), and hence show that the area of triangle \(A B C\) is 1.5 square units.  (2 marks)

  2. Find the shortest distance from point \(B\) to the line segment \(A C\).  (2 marks)

A second plane, \(\psi\), has the Cartesian equation  \(2 x-2 y-z=-18\).

  1. At what acute angle does the line given by \(\underset{\sim}{ r }(t)=3 \underset{\sim}{ i }+2 \underset{\sim}{ j }+4 \underset{\sim}{ k }+t(\underset{\sim}{ i }-2 \underset{\sim}{ j }+2\underset{\sim}{ k }), t \in R\), intersect the plane \(\psi\) ? Give your answer in degrees correct to the nearest degree.  (2 marks)

A line \(L\) passes through the origin and is normal to the plane \(\psi\). The line \(L\) intersects \(\psi\) at a point \(D\).

  1. Write down an equation of the line \(L\) in parametric form.  (1 mark)

  2. Find the shortest distance from the origin to the plane \(\psi\).  (2 marks)

  3. Find the coordinates of point \(D\).  (2 marks)

Show Answers Only

a.     \(\overrightarrow{A B}  =\overrightarrow{O B}-\overrightarrow{O A}=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)-\left(\begin{array}{l}1 \\ 1 \\ 2\end{array}\right)=\left(\begin{array}{l}0 \\ 1 \\ 1\end{array}\right)\)
  \(\overrightarrow{A C}  =\overrightarrow{O C}-\overrightarrow{O A}=\left(\begin{array}{l}3 \\ 2 \\ 4\end{array}\right)-\left(\begin{array}{l}1 \\ 1 \\ 2\end{array}\right)=\left(\begin{array}{l}2 \\ 1 \\ 2\end{array}\right)\)

\(\text{Area}=\dfrac{1}{2}\ \bigg|\overrightarrow{AB} \times \overrightarrow{AC}\bigg|=\dfrac{1}{2}\ \left|\begin{array}{ccc}\underset{\sim}{i} & \underset{\sim}{j} & \underset{\sim}{k} \\ 0 & 1 & 1 \\ 2 & 1 & 2\end{array}\right|=\dfrac{1}{2}\ \bigg|\underset{\sim}{i}+2 \underset{\sim}{j}-2 \underset{\sim}{k}\bigg|=\dfrac{3}{2}\)

b.    \(\text {Shortest distance }=1 \text { unit }\)

c.     \(26^{\circ}\)

d.    \(\underset{\sim}{r}(t)=\underset{\sim}{n} \times t=2 t \underset{\sim}{i}-2 t \underset{\sim}{j}-t \underset{\sim}{k}\)

\(x=2 t, y=-2 t, z=-t \quad \text{(also accepted)}\)

e.    \(\text {Shortest distance }=6\)

f.    \(D(-4,4,2)\)

Show Worked Solution

a.     \(\overrightarrow{A B}  =\overrightarrow{O B}-\overrightarrow{O A}=\left(\begin{array}{l}1 \\ 2 \\ 3\end{array}\right)-\left(\begin{array}{l}1 \\ 1 \\ 2\end{array}\right)=\left(\begin{array}{l}0 \\ 1 \\ 1\end{array}\right)\)
  \(\overrightarrow{A C}  =\overrightarrow{O C}-\overrightarrow{O A}=\left(\begin{array}{l}3 \\ 2 \\ 4\end{array}\right)-\left(\begin{array}{l}1 \\ 1 \\ 2\end{array}\right)=\left(\begin{array}{l}2 \\ 1 \\ 2\end{array}\right)\)

\(\text{Area}=\dfrac{1}{2}\ \bigg|\overrightarrow{AB} \times \overrightarrow{AC}\bigg|=\dfrac{1}{2}\ \left|\begin{array}{ccc}\underset{\sim}{i} & \underset{\sim}{j} & \underset{\sim}{k} \\ 0 & 1 & 1 \\ 2 & 1 & 2\end{array}\right|=\dfrac{1}{2}\ \bigg|\underset{\sim}{i}+2 \underset{\sim}{j}-2 \underset{\sim}{k}\bigg|=\dfrac{3}{2}\)

 
b.    \(\text {In } \triangle ABC \text {, shortest distance of } B \text { to } A C\ \text {is the} \perp \text {distance }\)

\(|\overrightarrow{A C}|= \displaystyle{\sqrt{2^2+1^2+2^2}}=3\)

\(\dfrac{3}{2}=\dfrac{1}{2}\ |\overrightarrow{A C}| \times h \ \Rightarrow \ h=1\)

\(\therefore \text { Shortest distance }=1 \text { unit }\)
 

♦♦ Mean mark (b) 35%.

c.    \(\text {Vector} \perp \text {to plane}\ \psi\  \text {is}\ \ \underset{\sim}{n}=2 \underset{\sim}{i}-2 \underset{\sim}{j}-\underset{\sim}{k}\)

\(\text{Parallel line to}\ \underset{\sim}{r}(t) \ \text{is}\ \ \underset{\sim}{m}=\underset{\sim}{i}-2 \underset{\sim}{j}+2\underset{\sim}{k}\)

\(\text{Find angle } \alpha \text{ between }\underset{\sim}{n} \text{ and } \underset{\sim}{m},\)

\(\text{Solve for } \alpha:\)

\((2 \underset{\sim}{i}-2 \underset{\sim}{j}-\underset{\sim}{k})(\underset{\sim}{i}-2 \underset{\sim}{j}+\underset{\sim}{2 k})=3 \times 3 \times \cos \, \alpha\)

\(\Rightarrow \alpha=64^{\circ} \text { (nearest degree) }\)
 

\(\text{Find angle } \theta \text{ between } \underset{\sim}{m} \text{ and plane } \psi :\)

\(\theta=90-64=26^{\circ}\)
 

d.    \(\underset{\sim}{r}(t)=\underset{\sim}{n} \times t=2 t \underset{\sim}{i}-2 t \underset{\sim}{j}-t \underset{\sim}{k}\)

\(x=2 t, y=-2 t, z=-t \quad \text{(also accepted)}\)
 

♦ Mean mark (d) 52%.

e.    \(|\overrightarrow{OD}|=\text{shortest distance from } O \text{ to plane } \psi\).

\(\Rightarrow D \text{ is on } L \text{ and } \psi\)

\(\text{Solve for } t: \  2(2 t)-2(-2 t)-(t)=-18\)

\(\Rightarrow t=-2\)

\(|\overrightarrow{OD}|=-4 \underset{\sim}{i}+4 \underset{\sim}{j}+2 \underset{\sim}{k}\)

\(\text {Shortest distance }=\displaystyle{\sqrt{(-4)^2+4^2+2^2}}=6\)
 

f.    \(\overrightarrow{OD}=-4 \underset{\sim}{i}+4 \underset{\sim}{j}+2 \underset{\sim}{i}\)

\(D(-4,4,2)\)

♦ Mean mark (f) 41%.

PHYSICS, M5 2021 VCE 3

To calculate the mass of distant pulsars, physicists use Newton's law of universal gravitation and the equations of circular motion.

The planet Phobetor orbits pulsar PSR B1257+12 at an orbital radius of 6.9 × 10\(^{10}\) m and with a period of 8.47 × 10\(^6\) s.

Assuming that Phobetor follows a circular orbit, calculate the mass of the pulsar. Show all your working.  (3 marks)

Show Answers Only

\(M=2.71\ \times 10^{30}\ \text{kg}\)

Show Worked Solution
\(\dfrac{r^3}{T^2}\) \(=\dfrac{GM}{4\pi^2}\)  
\(M\) \(=\dfrac{4\pi^2 r^3}{G T^2}\)  
  \(=\dfrac{4 \pi^2 \times (6.9 \times 10^{10})^3}{6.67 \times 10^{-11} \times (8.47 \times 10^6)^2}\)  
  \(=2.71 \times 10^{30}\ \text{kg}\)  

PHYSICS, M6 2021 VCE 2

A schematic side view of one design of an audio loudspeaker is shown in Figure 2. It uses a current carrying coil that interacts with permanent magnets to create sound by moving a cone in and out.
 

Figure 3 shows a schematic view of the loudspeaker from the position of the eye shown in Figure 2. The direction of the current is clockwise, as shown.
 

   

  1. Draw four magnetic field lines on Figure 3, showing the direction of each field line using an arrow.  (1 mark)

  2. Which one of the following gives the direction of the force acting on the current carrying coil shown in Figure 3?  (1 mark)
A. left B. right
C. up the page D. down the page
E. into the page F. out of the page
  1. The current carrying coil has a radius of 5.0 cm and 20 turns of wire, and it carries a clockwise current \((I)\) of 2.0 A. Its magnetic field strength \((B)\) is 200 mT.
  2. Calculate the magnitude of the force, \(F\), acting on the current carrying coil. Show your working.  (2 marks)

Show Answers Only

a.   
         

b.    \(E\)

c.    \(2.51\ \text{N}\)

Show Worked Solution

a.    Magnetic fields run from the north pole to the south pole.
 

   

 

b.    Apply the right-hand rule:

→ Thumb to the right and fingers down, the force on the current carrying coil must be into the page.

\(\Rightarrow E\)
 

c.    \(l= 2\pi \times 5 \times 10^{-2} = 0.1\pi\)

\(\therefore F=nlIB=20 \times 0.1\pi \times 2 \times 0.2=2.51\ \text{N}\)

♦♦♦ Mean mark 24%.

Statistics, SPEC2 2023 VCAA 6

A forest ranger wishes to investigate the mass of adult male koalas in a Victorian forest. A random sample of 20 such koalas has a sample mean of 11.39 kg.

It is known that the mass of adult male koalas in the forest is normally distributed with a standard deviation of 1 kg.

  1. Find a 95% confidence interval for the population mean (the mean mass of all adult male koalas in the forest). Give your values correct to two decimal places.  (1 mark)

  2. Sixty such random samples are taken and their confidence intervals are calculated.
  3. In how many of these confidence intervals would the actual mean mass of all adult male koalas in the forest be expected to lie?  (1 mark)

The ranger wants to decrease the width of the 95% confidence interval by 60% to get a better estimate of the population mean.

  1. How many adult male koalas should be sampled to achieve this?  (1 mark)

It is thought that the mean mass of adult male koalas in the forest is 12 kg. The ranger thinks that the true mean mass is less than this and decides to apply a one-tailed statistical test. A random sample of 40 adult male koalas is taken and the sample mean is found to be 11.6 kg.

  1. Write down the null hypothesis, \(H_0\), and the alternative hypothesis, \(H_1\), for the test.   (1 mark)

The ranger decides to apply the one-tailed test at the 1% level of significance and assumes the mass of adult male koalas in the forest is normally distributed with a mean of 12 kg and a standard deviation of 1 kg.

  1.  i. Find the \(p\) value for the test correct to four decimal places.  (1 mark)

  2. ii. Draw a conclusion about the null hypothesis in part d. from the \(p\) value found above, giving a reason for your conclusion.  (1 mark)

  3. What is the critical sample mean (the smallest sample mean for \(H_0\) not to be rejected) in this test? Give your answer in kilograms correct to three decimal places.  (1 mark)

Suppose that the true mean mass of adult male koalas in the forest is 11.4 kg, and the standard deviation is 1 kg. The level of significance of the test is still 1%.

  1. What is the probability, correct to three decimal places, of the ranger making a type \(\text{II}\) error in the statistical test?  (1 mark)

Show Answers Only

a.    \((10.95,11.83)\)

b.    \(57\)

c.    \(n=125\)

d.    \(H_0: \mu=12, \quad H_1: \mu<12\)

e.i.  \(p=0.0057\)

e.ii. \(\text{Since } p<0.01 \text { : reject } H_0 \text {, favour } H_1\)

f.    \(\text {Critical sample mean } \bar{x} \approx 11.632\)

g.    \(\text{Pr}(\bar{x} \geqslant 11.63217 \mid \mu=11.4) \approx 0.071\)

Show Worked Solution

a.    \(\sigma_{\text{pop}}=1\)

\(\text{Sample:}\ \ n=20,\ \ \bar{x}=11.39,\ \ \sigma_{\text {sample }}=\dfrac{1}{\sqrt{20}}\)

\(\text{Find 95% C.I. (by CAS):}\)

\((10.95,11.83)\)
 

b.    \(\text{95% C.I. for 60 samples calculated}\)

\(\text{Number expected }(\mu \text{ within C.I.)}=0.95 \times 60=57\)
 

c.    \(\text {C.I.}=\left(11.39-1.96 \times \dfrac{1}{\sqrt{20}}, 11.39+1.96 \times \dfrac{1}{\sqrt{20}}\right)\)

\(\Rightarrow \text { Interval }=2 \times 1.96 \times \dfrac{1}{\sqrt{20}}\)

\(\text{Interval reduced by } 60\%\)

\(\Rightarrow \text{ New interval }=0.40 \times 2 \times 1.96 \times \dfrac{1}{\sqrt{20}}\)

\(\text{Solve for } n\) :

\(2 \times 1.96 \times \dfrac{1}{\sqrt{n}}=0.40 \times 2 \times 1.96 \times \dfrac{1}{\sqrt{20}}\)

\(\Rightarrow n=125\)
 

♦♦♦ Mean mark (c) 28%.

d.    \(H_0: \mu=12,\ \ H_1: \mu<12\)
 

e.i.  \(E(\bar{X})=\mu=12\)

 \(\bar{x}=11.6, \ \sigma(\bar{X})=\dfrac{1}{\sqrt{40}}\)

 \(p=\operatorname{Pr}(\bar{X} \leqslant 11.6)=0.0057\)

 

e.ii. \(\text{Since}\ \  p<0.01 \text {: reject } H_0 \text {, favour } H_1\)
 

f.    \(\text{Pr}(\bar{X} \leqslant a \mid \mu=12) \geqslant 0.01\) 

\(\text{Find } a \text{ (by CAS):}\)

\(\text{inv Norm}\left(0.01,12, \dfrac{1}{\sqrt{40}}\right) \ \Rightarrow \ a \geqslant 11.63217\)

\(\text {Critical sample mean}\ \ \bar{x} \approx 11.632\)
 

g.    \(\mu=11.4,\ \ \sigma_{\text{pop}}=1,\ \  n=40\)

\(\text{Pr}(\bar{X} \geqslant 11.63217 \mid \mu=11.4) \approx 0.071\)

♦♦ Mean mark (g) 39%.

CHEMISTRY, M1 EQ-Bank 30

Explain the trend in reactivity with water of the elements in Group 2 as you move down the group.   (2 marks)

Show Answers Only

→ The reactivity of Group 2 elements with water increases as you move down the group due to decreasing ionization energies and increasing atomic radii.

→ Lower ionization energies and larger atomic radii make it easier for these metals to lose electrons and react with water, forming hydroxides and hydrogen gas. 

→ For instance, magnesium reacts slowly with hot water, while barium reacts vigorously even at room temperature, illustrating this trend.

Show Worked Solution

→ The reactivity of Group 2 elements with water increases as you move down the group due to decreasing ionization energies and increasing atomic radii.

→ Lower ionization energies and larger atomic radii make it easier for these metals to lose electrons and react with water, forming hydroxides and hydrogen gas. 

→ For instance, magnesium reacts slowly with hot water, while barium reacts vigorously even at room temperature, illustrating this trend.

PHYSICS, M8 2022 VCE 15

Figure 17 shows some of the energy levels of excited neon atoms. These energy levels are not drawn to scale.
 

  1. Show that the energy transition required for an emitted photon of wavelength 640 nm is 1.94 eV.  (1 mark)
  1. On Figure 17, draw an arrow to show the transition that would emit the photon described in part a.  (1 mark)

Show Answers Only

a.    \(E=hf=\dfrac{hc}{\lambda}=\dfrac{6.626 \times 10^{-34} \times 3 \times 10^8}{640 \times 10^{-9}}=3.106 \times 10^{-19}\ \text{J}\)

\(\text{Convert to electron volts} =\dfrac{3.106 \times 10^{-19}}{1.602 \times 10^{-19}}=1.94\ \text{eV}\)
 

b.   
       

Show Worked Solution

a.    \(E=hf=\dfrac{hc}{\lambda}=\dfrac{6.626 \times 10^{-34} \times 3 \times 10^8}{640 \times 10^{-9}}=3.106 \times 10^{-19}\ \text{J}\)

\(\text{Convert to electron volts} =\dfrac{3.106 \times 10^{-19}}{1.602 \times 10^{-19}}=1.94\ \text{eV}\)
 

b.
       

PHYSICS, M7 2022 VCE 14*

Sam undertakes a photoelectric effect experiment using the apparatus shown in Figure 12. She uses a green filter.
 

   

Sam produces a graph of photocurrent, \(I\), in milliamperes, versus voltage, \(V\), in volts, as shown in Figure 13.
 

   

  1. Identify what point \(\text{P}\) represents on the graph in Figure 13.  (1 mark)
  1. Sam then significantly increases the intensity of the light.
  2. Sketch the resulting graph on Figure 14. The dashed line in Figure 14 represents the original data.  (2 marks)
     

   

  1. Sam replaces the green filter with a violet filter, keeping the light source at the increased intensity.
  2. Sketch the resulting graph on Figure 15. The dashed line in Figure 15 represents the original data.  (2 marks)
     

Show Answers Only

a.    Stopping voltage

b.
      

c.    
       

Show Worked Solution

a.    Stopping voltage
 

b.    Diagram will exhibit the following points:

→ Increasing the intensity of light will increase the photocurrent due to more photoelectrons being ejected from the metal.

→ However, it will not change the energy of the photoelectrons, hence the stopping voltage remains the same.
 

c.    Diagram will exhibit the following points:

→ Changing the colour from green to violet will increase the frequency of the light.

→ As \(E=hf\), the energy of the photoelectrons will increase, hence there needs to be a greater stopping voltage.
 

Functions, MET2 2023 VCAA 2

The following diagram represents an observation wheel, with its centre at point \(P\). Passengers are seated in pods, which are carried around as the wheel turns. The wheel moves anticlockwise with constant speed and completes one full rotation every 30 minutes.When a pod is at the lowest point of the wheel (point \(A\)), it is 15 metres above the ground. The wheel has a radius of 60 metres.
 

Consider the function \(h(t)=-60\ \cos(bt)+c\) for some \(b, c \in R\), which models the height above the ground of a pod originally situated at point \(A\), after time \(t\) minutes.

  1. Show that \(b=\dfrac{\pi}{15}\) and \(c=75\).   (2 marks)
  2. Find the average height of a pod on the wheel as it travels from point \(A\) to point \(B\).
  3. Give your answer in metres, correct to two decimal places.   (2 marks)
  4. Find the average rate of change, in metres per minute, of the height of a pod on the wheel as it travels from point \(A\) to point \(B\).   (1 mark)

After 15 minutes, the wheel stops moving and remains stationary for 5 minutes. After this, it continues moving at double its previous speed for another 7.5 minutes.

The height above the ground of a pod that was initially at point \(A\), after \(t\) minutes, can be modelled by the piecewise function \(w\):
 

\(w(t) = \begin {cases}
h(t)         &\ \ 0 \leq t < 15 \\
k         &\ \ 15 \leq t < 20 \\
h(mt+n) &\ \ 20\leq t\leq 27.5
\end{cases}\)

 
where \(k\geq 0, m\geq 0\) and \(n \in R\).

  1.   i.State the values of \(k\) and \(m\).   (1 mark)
       
     ii. Find all possible values of \(n\).   (2 marks)

      

    iii. Sketch the graph of the piecewise function \(w\) on the axes below, showing the coordinates of the endpoints.   (3 marks)


     
  1.  
Show Answers Only

a.    \(\text{See worked solution}\)

b.    \(\approx 36.80\ \text{m}\)

c.    \(8\)

d.i.  \(k=135, m=2\)

d.ii. \(n=30p+5,\ p\in Z\)

d.iii. 

Show Worked Solution
a.    \(\text{Period:}\) \(\dfrac{2\pi}{b}\) \(=30\)
    \(\therefore\ b\) \(=\dfrac{\pi}{15}\)

  
\(\text{Given }h(t)=-60\cos(bt)+c,\ \text{evaluate when }t=0, h=15\ \text{to find }c.\)

\(-60\ \cos(0)+c\)  \(=15\)
\(c\) \(=75\)

  \(\therefore\ h(t)=-60\cos(\dfrac{\pi t}{15})+75\)

b.     \(\text{Average height}\) \(=\dfrac{1}{\frac{30}{4}-0}\displaystyle\int_0^{\frac{30}{4}}\Bigg(-60\cos\Bigg(\dfrac{\pi}{15}t\Bigg)+75\Bigg)\,dt\)
    \(=\dfrac{2}{15}\left[75t-\dfrac{900\ \sin(\frac{\pi}{15}t)}{\pi}\right]_{0}^{7.5}\)
    \(=\dfrac{2}{15}\Bigg[75\times 7.5-\dfrac{900\ \sin(\frac{\pi}{15}\times 7.5)}{\pi}\Bigg]-\Bigg[0\Bigg]\)
    \(=\dfrac{75\pi-120}{\pi}=\dfrac{15(5\pi-8)}{\pi}\)
    \(=36.802\dots\approx 36.80\ \text{m}\)
 
♦♦ Mean mark (b) 45%.
MARKER’S COMMENT: \(\frac{1}{60}\int_0^{60} h(t)dt\) was a common error.
Others incorrectly found average rate of change instead of average value.

c.   \(\text{Av rate of change of height}\)

  \(=\dfrac{h(7.5)-h(0)}{7.5}\)
  \(=\dfrac{\Bigg(75-60\cos(\dfrac{\pi \times 7.5}{15})\Bigg)-\Bigg(75-60\cos(\dfrac{\pi\times 0}{15})\Bigg)}{7.5}\)
  \(=\dfrac{75-15}{7.5}=8\)
 
♦ Mean mark (c) 50%.
MARKER’S COMMENT: Common incorrect answer was – 8.
di.    \(\text{Period is 30 minutes, so after 15 minutes pod}\)
\(\text{is at the top of the wheel.}\)
  \(\therefore\ k=75-60\ \cos(\frac{\pi}{15}\times 15)=135\)

  
\(\text{Pod is travelling at twice its previous speed}\)
\(\text{so one revolution takes 15 minutes}\)

\(\therefore\ \text{Period}\rightarrow\) \(\dfrac{2\pi}{bm}\) \(=15\)
  \(bm\) \(=\dfrac{2\pi}{15}\)
  \(\dfrac{\pi}{15}\cdot m\) \(=\dfrac{2\pi}{15}\)
  \(m\) \(=\dfrac{2\pi}{15}\times \dfrac{15}{\pi}\)
    \(=2\)
 
♦♦ Mean mark (d)(i) 40%.
MARKER’S COMMENT: Many students could find \(k\) but not \(m\) with \(m=\frac{1}{2}\) a common error.
dii.   \(\text{When }t=20\ \text{the pod is at the top of the wheel and height is 135.}\)
  
\(\text{When }t=27.5\ \text{the pod is back at the start and height is 15.}\)

   
\(\text{Using CAS solve for }t=20:\rightarrow\ h(2(20)+n)=135\ \rightarrow\ n=30p+5\)

\(\text{Using CAS solve for }t=27.5:\rightarrow\ h(2(27.5)+n)=15\ \rightarrow\ n=30p+5\)

\(\therefore\ n=30p+5,\ p\in Z\)

 
♦♦♦ Mean mark (d)(ii) 20%.
MARKER’S COMMENT: Many students set up the correct equations to solve but did not provide the general solution. Some incorrectly stated the variable as an element of R.

diii.   

 
♦♦♦ Mean mark (d)(iii) 40%.
MARKER’S COMMENT: Students are reminded to include all endpoints and be particular about the curvature of graph.

Calculus, MET2 2023 VCAA 1

Let \(f:R \rightarrow R, f(x)=x(x-2)(x+1)\). Part of the graph of \(f\) is shown below.

  1. State the coordinates of all axial intercepts of \(f\).   (1 mark)
  2. Find the coordinates of the stationary points of \(f\).   (2 marks)
    1. Let \(g:R\rightarrow R, g(x)=x-2\).
    2. Find the values of \(x\) for which \(f(x)=g(x)\).   (1 mark)
    1. Write down an expression using definite integrals that gives the area of the regions bound by \(f\) and \(g\).  (2 marks)
    2. Hence, find the total area of the regions bound by \(f\) and \(g\), correct to two decimal places.   (1 mark)
  1. Let \(h:R\rightarrow R, h(x)=(x-a)(x-b)^2\), where \(h(x)=f(x)+k\) and \(a, b, k \in R\).
  2. Find the possible values of \(a\) and \(b\).   (4 marks)
Show Answers Only

a.    \((-1, 0), (0, 0), (2, 0)\)

b.    \(\Bigg(\dfrac{1-\sqrt{7}}{3}, \dfrac{2(7\sqrt{7}-10}{27}\Bigg), \Bigg(\dfrac{1+\sqrt{7}}{3}, \dfrac{-2(7\sqrt{7}-10}{27}\Bigg)\)

c.i.  \(x=2,\ \text{or}\ x=\dfrac{-1\pm \sqrt{5}}{2}\)

c.ii. \(\text{Let }a=\dfrac{-1-\sqrt{5}}{2}\ \text{and }b=\dfrac{-1+\sqrt{5}}{2}\)

 \(\text{Then}\ A=\displaystyle\int_a^b (x-2)(x^2+x-1)\,dx+\displaystyle\int_b^2 -(x-2)(x^2+x-1)\,dx\)

c.iii. \(5.95\)

d.   \(\text{1st case }\rightarrow \ a=\dfrac{2\sqrt{7}+1}{3}, b=\dfrac{1-\sqrt{7}}{3}\)

\(\text{2nd case }\rightarrow \ a=\dfrac{-2\sqrt{7}+1}{3}, b=\dfrac{1+\sqrt{7}}{3}\)

Show Worked Solution

a.    \((-1, 0), (0, 0), (2, 0)\)
  

b.    \(\text{Using CAS solve for}\ x:\)

\(\dfrac{d}{dx}(x(x-2)(x+1))=0\)

\(\therefore\ x=\dfrac{1-\sqrt{7}}{3}\ \text{and }x=\dfrac{1+\sqrt{7}}{3}\)

\(\text{Substitute }x\ \text{values into }f(x)\ \text{using CAS to get}\ y\ \text{values}\)

\(\text{The stationary points of }f\ \text{are}:\)

\(\Bigg(\dfrac{1-\sqrt{7}}{3}, \dfrac{2(7\sqrt{7}-10}{27}\Bigg), \Bigg(\dfrac{1+\sqrt{7}}{3}, \dfrac{-2(7\sqrt{7}-10}{27}\Bigg)\)
  

ci    \(\text{Given }f(x)=g(x)\)

\(x(x-2)(x+1)\) \(=x-2\)
\(x(x-2)(x+1)(x-2)\) \(=0\)
\((x-2)(x(x+1)-1)\) \(=0\)
\((x-2)(x^2+x-1)\) \(=0\)

  
\(\therefore\ \text{Using CAS: } \)

\(x=2,\ \text{or}\ x=\dfrac{-1\pm \sqrt{5}}{2}\)

cii  \(\text{Area of bounded region:}\)

\(\text{Let }a=\dfrac{-1-\sqrt{5}}{2}\ \text{and }b=\dfrac{-1+\sqrt{5}}{2}\)

\(\text{Then}\ A=\displaystyle\int_a^b (x-2)(x^2+x-1)\,dx+\displaystyle\int_b^2 -(x-2)(x^2+x-1)\,dx\)
  

ciii  \(\text{Solve the integral in c.ii above using CAS:}\)
  \(\text{Total area}=5.946045..\approx 5.95\)

  

d.   \(\text{Method 1 – Equating coefficients}\)

\((x-a)(x-b)^2=x(x-2)(x+1)+k\)

\(x^3-2bx^2-ax^2+b^2x+2abx-ab^2=x^3-x^2-2x+k\)

\((x^3-(a+2b)x^2+(2ab+b^2)x-ab^2=x^3-x^2-2x+k\)

\(\therefore\ -(a+2b)=-1\ \to\ a=1-2b …(1)\)

\(2ab+b^2=-2\ \ …(2)\)

\(\text{Substitute (1) into (2) and solve for }b.\)

\(2b(1-2b)+b^2\) \(=-2\)
\(3b^2-2b-2\) \(=0\)
\(b\) \(=\dfrac{1\pm \sqrt{7}}{3}\)
\(\text{When }b\) \(=\dfrac{1+\sqrt{7}}{3}\)
\(a\) \(=1-2\Bigg(\dfrac{1+\sqrt{7}}{3}\Bigg)=\dfrac{-2\sqrt{7}+1}{3}\)
\(\text{When }b\) \(=\dfrac{1-\sqrt{7}}{3}\)
\(a\) \(=1-2\Bigg(\dfrac{1-\sqrt{7}}{3}\Bigg)=\dfrac{2\sqrt{7}+1}{3}\)

  

\(\text{Method 2 – Using transformations}\)

\(\text{The squared factor in }(x-a)(x-b)^2=x(x-2)(x+1)+k,\)

\(\text{shows that the turning point is on the }x\ \text{axis}.\)

\(\therefore\ \text{Lowering }f(x)\ \text{by }\dfrac{2(7\sqrt{7}-10)}{27}\ \text{and raising }f(x)\ \text{by }\dfrac{2(7\sqrt{7}+10)}{27}\)

\(\text{will give the 2 possible sets of values for }a\ \text{and}\ b.\)

\(\text{1st case – lowering using CAS solve }h(x) =0\ \rightarrow\ h(x)=f(x)-\dfrac{2(7\sqrt{7}-10)}{27}\)

\(\therefore\ x-\text{intercepts}\rightarrow \ a=\dfrac{2\sqrt{7}+1}{3}, b=\dfrac{1-\sqrt{7}}{3}\)

\(\text{2nd case – raising using CAS solve }h(x) =0\ \rightarrow\ h(x)=f(x)+\dfrac{2(7\sqrt{7}+10)}{27}\)

\(\therefore\ x-\text{intercepts}\rightarrow \ a=\dfrac{-2\sqrt{7}+1}{3}, b=\dfrac{1+\sqrt{7}}{3}\)

 

PHYSICS, M8 2021 VCE 17 MC

Which one of the following is closest to the de Broglie wavelength of a 663 kg motor car moving at 10 m s\(^{-1}\) ?

  1. \(10^{-37}\) m
  2. \(10^{-36}\) m
  3. \(10^{-35}\) m
  4. \(10^{-34}\) m
Show Answers Only

\(A\)

Show Worked Solution

\(\lambda=\dfrac{h}{mv}=\dfrac{6.626 \times 10^{-34}}{663 \times 10}=9.99 \times 10^{-38} \approx 10^{-37}\ \text{m}\)

\(\Rightarrow A\)

PHYSICS, M5 2021 VCE 9-10 MC

Lucy is running horizontally at a speed of 6 m s\(^{-1}\) along a diving platform that is 8.0 m vertically above the water.

Lucy runs off the end of the diving platform and reaches the water below after time \(t\).

She lands feet first at a horizontal distance \(d\) from the end of the diving platform.
 

Question 9

Which one of the following expressions correctly gives the distance \(d\) ?

  1. 0.8\(t\)
  2. 6\(t\)
  3. 5\(t^2\)
  4. 6\(t\) + 5\(t^2\)

 
Question 10

Which one of the following is closest to the time taken, \(t\), for Lucy to reach the water below?

  1. 0.8 s
  2. 1.1 s
  3. 1.3 s
  4. 1.6 s
Show Answers Only

\(\text{Question 9:} \ B\)

\(\text{Question 10:} \ C\)

Show Worked Solution

\(\text{Question 9}\)

→ The horizontal displacement formula for projectile motion:  \(s=ut\)

→ Horizontal distance: \(d=6t\)

\(\Rightarrow B\)
 

\(\text{Question 10}\)

\(\text{Find time of flight}\ (t):\)

\(s\) \(=ut+\dfrac{1}{2}at^2\)  
\(s\) \(=\dfrac{1}{2}at^2\ \ \ (u=0)\)  
\(t\) \(=\sqrt{\dfrac{2s}{a}}\)  
  \(=\sqrt{\dfrac{2 \times 8}{9.8}}\)  
  \(\approx 1.3\ \text{s}\)  

 

\(\Rightarrow C\)

PHYSICS, M6 2021 VCE 7*

A mobile phone charger uses a step-down transformer to transform 240 V AC mains voltage to 5.0 V. The mobile phone draws a current of 3.0 A while charging. Assume that the transformer is ideal and that all readings are RMS.

Calculate the current drawn from the mains during charging?   (2 marks)

Show Answers Only

\(0.06\ \text{A}\)

Show Worked Solution
\(\text{Power}_{\text{in}}\) \(=\ \text{Power}_{\text{out}}\)  
\(V_p I_p\) \(=V_s I_s\)  
\(I_p\) \(=\dfrac{V_s I_s}{V_p}\)  
  \(=\dfrac{5 \times 3}{240}\)  
  \(=0.06\ \text{A}\)  

PHYSICS, M6 2021 VCE 6 MC

A magnet approaches a coil with six turns, as shown in the diagram below. During time interval \(\Delta t\), the magnetic flux changes by 0.05 Wb and the average induced EMF is 1.2 V.
 

Which one of the following is closest to the time interval \(\Delta t\) ?

  1. 0.04 s
  2. 0.01 s
  3. 0.25 s
  4. 0.50 s
Show Answers Only

\(C\)

Show Worked Solution
\(\varepsilon\) \(=N\dfrac{\Delta \Phi}{\Delta t}\)  
\(\Delta t\) \(=N \dfrac{\Delta \Phi}{\varepsilon}\)  
  \(=6 \times \dfrac{0.05}{1.2}\)  
  \(=0.25\ \text{s}\)  

 

\(\Rightarrow C\)

PHYSICS, M8 2022 VCE 17

A materials scientist is studying the diffraction of electrons through a thin metal foil. She uses electrons with an energy of 10.0 keV. The resulting diffraction pattern is shown in Figure 19.
 

  1. Calculate the de Broglie wavelength of the electrons in nanometres.  (4 marks)

  1. The materials scientist then increases the energy of the electrons by a small amount and hence their speed by a small amount.

    Explain what effect this would have on the de Broglie wavelength of the electrons. Justify your answer.  (3 marks)

Show Answers Only

a.    \(0.012\ \text{nm}\)

 

b.    Effects when speed of the electron increases by a small amount:

→ Momentum will increase by a small amount

→ As \(\lambda \propto \dfrac{1}{mv}\), if the momentum of the electron increases, its corresponding de Broglie wavelength will decrease.

Show Worked Solution

a.    Convert electron volts to joules:

\(\Rightarrow \ E=10 \times 10^3 \times 1.602 \times 10^{-19}=1.602 \times 10^{-15}\ \text{J}\)

\(E\) \(=\dfrac{1}{2}mv^2\)  
\(v\) \(=\sqrt{\dfrac{2E}{m}}\)  
  \(=\sqrt{\dfrac{2 \times 1.602 \times 10^{-15}}{9.109 \times 10^{-31}}}\)  
  \(=5.93 \times 10^7\)  

 

\(\therefore \lambda\) \(=\dfrac{h}{mv}\)  
  \(=\dfrac{6.626 \times 10^{-34}}{9.109 \times 10^{-27} \times 5.93 \times 10^7}\)  
  \(=1.23 \times 10^{-11}\ \text{m}\)  
  \(=0.012\ \text{nm}\)  

♦ Mean mark 42%.

b.    Effects when speed of the electron increases by a small amount:

→ Momentum will increase by a small amount

→ As \(\lambda \propto \dfrac{1}{mv}\), if the momentum of the electron increases, its corresponding de Broglie wavelength will decrease.

Calculus, 2ADV C4 2022 MET1 4

The graph of  \(y=x+\dfrac{1}{x}\) is shown over part of its domain.
 

Use two trapeziums of equal width to approximate the area between the curve, the \(x\)-axis and the lines  \(x=1\)  and  \(x=3\).   (2 marks)

Show Answers Only

\(5\dfrac{1}{6}\)

Show Worked Solution

\(\text{Trapezium rule approximation (see formula sheet):}\)

\begin{array} {|c|c|}
\hline
\rule{0pt}{2.5ex} x\rule[-1ex]{0pt}{0pt} & 1&2&3 \\
\hline
\rule{0pt}{2.5ex} f(x)\rule[-1ex]{0pt}{0pt} & 1+1=2 & 2+\dfrac{1}{2}=\dfrac{5}{2} & 3+ \dfrac{1}{3}=\dfrac{10}{3}\\
\hline
\end{array}

\(\text{Area}\) \(\approx \dfrac{3-1}{2\times 2}\Bigg[2+2\times\dfrac{5}{2}+\dfrac{10}{3}\Bigg]\)
  \(\approx\dfrac{1}{2}\Bigg[\dfrac{6}{3}+\dfrac{15}{3}+\dfrac{10}{3}\Bigg]\)
  \(\approx5\dfrac{1}{6}\)

Calculus, 2ADV C2 2023 MET1 1a

Let  \(y=\dfrac{x^2-x}{e^x}\).

Find and simplify \(\dfrac{dy}{dx}\).   (2 marks)

Show Answers Only

\(\dfrac{-x^2+3x-1}{e^x}\)

Show Worked Solution

\(\text{Using the quotient rule:}\)

\(\dfrac{dy}{dx}\) \(=\dfrac{e^x(2x-1)-(x^2-x)e^x}{(e^x)^2}\)
  \(=\dfrac{e^x(-x^2+3x-1)}{e^{2x}}\)
  \(=\dfrac{-x^2+3x-1}{e^x}\)

L&E, 2ADV E1 2023 MET1 2

Solve  \(e^{2x}-12=4e^{x}\)  for  \(x\ \in\ R\).   (3 marks)

Show Answers Only

\(x=\log_{e}6\)

Show Worked Solution
\(e^{2x}-12\) \(=4e^{x}\)
\(e^{2x}-4e^{x}-12\) \(=0\)

 
\(\text{Let}\ \ u=e^{x}:\)

\(u^2-4u-12\) \(=0\)
\((u-6)(u+2)\) \(=0\)
 

\(\Rightarrow u=6\ \ \ \text{or}\ -2\)

\(\therefore e^{x}\) \(=6\ \ \ \ \ \ \ \ \ \ \text{or}\ \ \ \ \ \ e^{x}=-2\ \text{(no solution)}\)  
\(x\) \(=\log_{e}6 \)  

PHYSICS, M6 2022 VCE 6

Figure 6 shows a simple alternator consisting of a rectangular coil of area 0.060 m\(^{2}\) and 200 turns, rotating in a uniform magnetic field. The magnetic flux through the coil in the vertical position shown in Figure 6 is 1.2 × 10\(^{-3}\) Wb.
 

  1. Calculate the strength of the magnetic field in Figure 6 . Show your working.  (2 marks)

  1. The rectangular coil rotates at a frequency of 2.5 Hz.
  2. Calculate the average induced EMF produced in the first quarter of a turn. Begin the quarter with the coil in the vertical position shown in Figure 6.  (3 marks)

Show Answers Only

a.    \(0.02\ \text{T}\)

b.    \(2.4\ \text{V}\)

Show Worked Solution
a.     \(\Phi\) \(=BA\)
  \(B\) \(=\dfrac{\Phi}{A}\)
    \(=\dfrac{1.2 \times 10^{-3}}{0.060}\)
    \(=0.02\ \text{T}\)

 
b.    \(\text{The time for 1 complete rotation:}\)

\(T=\dfrac{1}{f}=\dfrac{1}{2.5}=0.4\ \text{s}\)

\(\Rightarrow \text{Time for a quarter turn}\ =0.1\ \text{s}\)
 
\(\varepsilon\) \(=N\dfrac{\Delta \Phi}{t}\)  
  \(=200 \times  \dfrac{1.2 \times 10^{-3}}{0.1}\)  
  \(=2.4\ \text{V}\)  

PHYSICS, M6 2022 VCE 5*

A wind generator provides power to a factory located 2.00 km away, as shown in Figure 5.

When there is a moderate wind blowing steadily, the generator produces a voltage of 415 V and a current of 100 A.

The total resistance of the transmission wires between the wind generator and the factory is 2.00 \(\Omega\).
 

  1. Calculate the power, in kilowatts, produced by the wind generator when there is a moderate wind blowing steadily.  (1 mark)

To operate correctly, the factory's machinery requires a power supply of 40 kW.

  1. Determine whether the energy supply system, as shown, will be able to supply power to the factory when the moderate wind is blowing steadily. Justify your answer with calculations.  (3 marks)

  1. The factory's owner decides to limit transmission energy loss by installing two transformers: a step-up transformer with a turns ratio of 1:10 at the wind generator and a step-down transformer with a turns ratio of 10:1 at the factory. Each transformer can be considered ideal.

    With the installation of the transformers, determine the power, in kilowatts, now supplied to the factory.  (3 marks)

Show Answers Only

a.    \(41.5\ \text{kW}\)

b.    \(P_{\text{loss}}=I^2R=100^2 \times 2=20000\ \text{W}=20\ \text{kW}\)

\(\Rightarrow\ \text{Net power supplied}\ =41.5-20=21.5\ \text{kW}\ < 40\ \text{kW}\)

\(\therefore \ \text{The power supply will not be enough to power the factory.}\)

c.    \(41.3\ \text{kW}\)

Show Worked Solution

a.    \(P=VI=415 \times 100 = 41\ 500\ \text{W}=41.5\ \text{kW}\)
 

b.    \(P_{\text{loss}}=I^2R=100^2 \times 2=20000\ \text{W}=20\ \text{kW}\)

\(\Rightarrow\ \text{Net power supplied}\ =41.5-20=21.5\ \text{kW}\ < 40\ \text{kW}\)

\(\therefore \ \text{The power supply will not be enough to power the factory.}\)
 

c.     \(\dfrac{I_s}{I_p}\) \(=\dfrac{N_p}{N_s}\)
  \(I_s\) \(=\dfrac{N_p}{N_s} \times I_p\)
    \(=\dfrac{1}{10} \times 100\)
    \(=10\ \text{A}\)

 

\(P_{\text{loss}} \text{(new)} =I^2R=10^2 \times 2=200\ \text{W}=0.2\ \text{kW}\)

\(\Rightarrow\ \text{Net power supplied}\ =41.5-0.2=41.3\ \text{kW}\)

PHYSICS, M6 2022 VCE 3

A schematic diagram of a mass spectrometer that is used to deflect charged particles to determine their mass is shown in Figure 3. Positive singly charged ions (with a charge of +1.602 × 10\(^{-19}\) C) are produced at the ion source. These are accelerated between an anode and a cathode. The potential difference between the anode and the cathode is 1500 V. The ions pass into a region of uniform magnetic field, \(B\), and are directed by the field into a semicircular path of diameter \(D\).
 

   

  1. Calculate the increase in the kinetic energy of each ion as it passes between the anode and the cathode. Give your answer in joules.  (2 marks)

Each ion has a mass of 4.80 × 10\(^{-27}\) kg.

  1. Show that each ion has a speed of 3.16 × 10\(^{5}\) m s\(^{-1}\) when it exits the cathode. Assume that the ion leaves the ion source with negligible speed. Show your working.  (2 marks)
  1. The region of uniform magnetic field, \(B\), in Figure 3 has a magnitude of 0.10 T.
  2. Calculate the diameter, \(D\), of the semicircular path followed by the ions within the magnetic field in Figure 3.  (3 marks)
Show Answers Only

a.    \(2.403 \times 10^{-16}\ \text{J}\)

b.     \(\Delta KE\) \(=\dfrac{1}{2}mv^2\)
  \(v\) \(=\sqrt{\dfrac{2 \Delta KE}{m}}\)
    \(=\sqrt{\dfrac{2 \times 2.403 \times 10^{-16}}{4.80 \times 10^{-27}}}\)
    \(=3.16 \times 10^5\ \text{ms}^{-1}\)

c.    \(0.19\ \text{m}\)

Show Worked Solution
a.    \(W=\Delta KE=qV=1.602 \times 10^{-19} \times 1500=2.403 \times 10^{-16}\ \text{J}\)
  
b.     \(\Delta KE\) \(=\dfrac{1}{2}mv^2\)
  \(v\) \(=\sqrt{\dfrac{2 \Delta KE}{m}}\)
    \(=\sqrt{\dfrac{2 \times 2.403 \times 10^{-16}}{4.80 \times 10^{-27}}}\)
    \(=3.16 \times 10^5\ \text{ms}^{-1}\)

 

c.     \(F_B\) \(=F_c\)
  \(qvB\) \(=\dfrac{mv^2}{r}\)
  \(r\) \(=\dfrac{mv}{qB}\)
    \(=\dfrac{4.80 \times 10^{-27} \times 3.16 \times 10^5}{1.602 \times 10^{-19} \times 0.10}\)
    \(=0.095\ \text{m}\)
  \(\therefore D\) \(=0.19\ \text{m}\)

PHYSICS, M5 2022 VCE 2

There are over 400 geostationary satellites above Earth in circular orbits. The period of orbit is one day (86 400 seconds). Each geostationary satellite remains stationary in relation to a fixed point on the equator. Figure 2 shows an example of a geostationary satellite that is in orbit relative to a fixed point, \(\text{X}\), on the equator.
 

  1. Explain why geostationary satellites must be vertically above the equator to remain stationary relative to Earth's surface.  (2 marks)
  1.  Using  \(G=6.67 \times 10^{-11}\ \text{N m}^{2}\ \text{kg}^{-2}, M_{\text{E}} = 5.98 \times 10^{24}\ \text{kg}\)  and  \(R_{\text{E}} = 6.37 \times 10^{6}\ \text{m}\), show that the altitude of a geostationary satellite must be equal to \(3.59 \times 10^{7}\ \text{m}\).  (4 marks)
  1. Calculate the speed of an orbiting geostationary satellite.  (3 marks)

Show Answers Only

a.   Reasons the satellite must orbit relative to a fixed point on the equator:

→ So that it is in the same rotational plane/axis of the Earth. 

→ The gravitational force on the satellite must be directly towards the centre of the Earth and it must orbit at a set altitude to the Earth where the speed of the satellite matches the rotational speed of the Earth.

→ As the speed and direction of the satellite matches that of the Earth, it will remain stationary relative to the motion of the Earth.

b.    See worked solutions

c.    3076 ms\(^{-1}\)

Show Worked Solution

a.   Reasons the satellite must orbit relative to a fixed point on the equator:

→ So that it is in the same rotational plane/axis of the Earth. 

→ The gravitational force on the satellite must be directly towards the centre of the Earth and it must orbit at a set altitude to the Earth where the speed of the satellite matches the rotational speed of the Earth.

→ As the speed and direction of the satellite matches that of the Earth, it will remain stationary relative to the motion of the Earth.
 

♦♦♦ Mean mark (a) 19%.
b.     \(\dfrac{r^3}{T^2}\) \(=\dfrac{GM}{4\pi^2}\)
  \(r\) \(=\sqrt[3]{\dfrac{GMT^2}{4\pi^2}}\)
    \(=\sqrt[3]{\dfrac{6.67 \times 10^{-11} \times 5.98 \times 10^{24} \times (86\ 400)^2}{4\pi^2}}\)
    \(=42.250 \times 10^6\ \text{m}\)

 
\(\therefore \ \text{Altitude}\ =42.250 \times 10^6 -6.37 \times 10^6 =3.59 \times 10^7\ \text{m}\)
 

c.     \(v\) \(=\sqrt{\dfrac{GM}{r}}\)
    \(=\sqrt{\dfrac{6.67 \times 10^{-11} \times 5.98 \times 10^{24}}{42.25 \times 10^6}}\)
    \(=3073\ \text{ms}^{-1}\)

PHYSICS, M6 2022 VCE 1

Figure 1 shows four positions (1, 2, 3 and 4) of the coil of a single-turn, simple DC motor. The coil is turning in a uniform magnetic field that is parallel to the plane of the coil when the coil is in Position 1, as shown.

When the motor is operating, the coil rotates about the axis through the middle of sides \(L M\) and \(N K\) in the direction indicated. The coil is attached to a commutator. Current for the motor is passed to the commutator by brushes that are not shown in Figure 1.
 

  1. When the coil is in Position 1, in which direction is the current flowing in the side \(K L-\) from \(K\) to \(L\) or from \(L\) to \(K\)? Justify your answer.  (2 marks)

  1. When the coil is in Position 3, in which direction is the current flowing in the side \(KL-\) from \(K\) to \(L\) or
    from \(L\) to \(K\)?      (1 marks)

  1. The side \(K L\) of the coil has a length of 0.10 m and experiences a magnetic force of 0.15 N due to the magnetic field, which has a magnitude of 0.5 T.

    Calculate the magnitude of the current in the coil.  (2 marks)

Show Answers Only

a.    \(K\) to \(L\)

b.    \(L\) to \(K\)

c.    \(3\ \text{A}\)

Show Worked Solution

a.    Using the right hand rule:

→ Palm faces up (force), fingers to the right (magnetic field), and the thumb faces out of the page.

→ Therefore the current must run from \(K\) to \(L\).
 

b.    Using the right hand rule:

→ Palm now faces down the page (force), fingers still to the right.

→ The current will run from \(L\) to \(K\) (thumb).

 

c.    \(F\) \(=lIB\)
  \(I\) \(=\dfrac{F}{lB}\)
    \(=\dfrac{0.15}{0.1 \times 0.5}\)
    \(=3\ \text{A}\)

PHYSICS, M8 2022 VCE 17 MC

Gamma radiation is often used to treat cancerous tumours. The energy of a gamma photon emitted by radioactive cobalt-60 is 1.33 MeV.

Which one of the following is closest to the frequency of the gamma radiation?

  1. \(1.33 \times 10^{6}\ \text{Hz}\)
  2. \(3.21 \times 10^{20}\ \text{Hz}\)
  3. \(3.21 \times 10^{21}\ \text{Hz}\)
  4. \(2.01 \times 10^{39}\ \text{Hz}\)
Show Answers Only

\(B\)

Show Worked Solution

→ \(\text{Convert 1.33 MeV to joules:}\)

\(1.33 \times 10^{6} \times 1.602 \times 10^{-19} =2.13 \times 10^{-13}\ \text{J}\)

→ \(E=hf \ \Rightarrow\ \ f=\dfrac{E}{h} = \dfrac{2.13 \times 10^{-13}}{6.626 \times 10^{-34}}=3.21 \times 10^{20}\ \text{Hz}\)

\(\Rightarrow B\)

PHYSICS, M8 2022 VCE 14 MC

Which one of the following best provides evidence of electrons behaving as waves?

  1. photoelectric effect
  2. atomic emission spectra
  3. atomic absorption spectra
  4. diffraction of electrons through a crystal
Show Answers Only

\(D\)

Show Worked Solution

→ Diffraction is a wave phenomenon.

→ As electrons produce a diffraction pattern after passing through a crystal, they are behaving as waves.

\(\Rightarrow D\)

PHYSICS, M6 2022 VCE 5 MC

A simple electricity generator is shown in the diagram below. When the coil is rotated, the output voltage across the slip rings is measured. The graph shows how the output voltage varies with time.
 

The frequency of rotation of the generator is now doubled.

Which one of the following graphs best represents the output voltage measured across the slip rings?
 

Show Answers Only

\(C\)

Show Worked Solution

→ If the frequency is doubled than the coil will rotate twice as fast, hence two periods will occur instead of one.

→ The induced emf is proportional to the frequency of the rotation of the coil, as the frequency is doubled so will the induced emf.

\(\Rightarrow C\)

PHYSICS, M8 2023 VCE 16

Fluorescent lights, when operating, contain gaseous mercury atoms, as shown in Figure 17.
 

Analysis of the light produced by fluorescent lights shows a number of emission spectral lines, including a prominent line representing a wavelength of 436.6 nm.

  1. Calculate the energy of the photons represented by the emission spectral line representing a wavelength of 436.6 nm.  (2 marks)

Figure 18 shows the lowest five energy levels for mercury.
 

  1. On the energy level diagram in Figure 18, draw an arrow showing the energy level transition that corresponds to the production of the spectral line representing a wavelength of 436.6 nm.  (1 mark)

Show Answers Only

a.    \(E=2.84\ \text{eV}\)

b.    
       

Show Worked Solution

a.    \(E=\dfrac{hc}{\lambda}=\dfrac{6.626 \times 10^{-34} \times 3 \times 10^8}{436.6 \times 10^{-9}}=4.549 \times 10^{-19}\ \text{J}\)

\(\text{Convert to eV:}\)

\(E= \dfrac{4.549 \times 10^{-19}}{1.602 \times 10^{-19}} = 2.84\ \text{eV}\)

 

b.    

Calculus, SPEC2 2023 VCAA 4

A fish farmer releases 200 fish into a pond that originally contained no fish. The fish population, \(P\), grows according to the logistic model,  \(\dfrac{d P}{d t}=P\left(1-\dfrac{P}{1000}\right)\) , where \(t\) is the time in years after the release of the 200 fish.

  1. The above logistic differential equation can be expressed as
  2. \(\displaystyle \int \frac{A}{P}+\dfrac{B}{1-\dfrac{P}{1000}} d P=\int dt \text {, where } A, B \in R .\)
  3. Find the values of \(A\) and \(B\).  (1 mark)

One form of the solution for \(P\) is  \(P=\dfrac{1000}{1+D e^{-t}}\ \),  where \(D\) is a real constant.

  1. Find the value of \(D\).  (1 mark)

The farmer releases a batch of \(n\) fish into a second pond, pond 2 , which originally contained no fish. The population, \(Q\), of fish in pond 2 can be modelled by  \(Q=\dfrac{1000}{1+9 e^{-1.1 t}}\),  where \(t\) is the time in years after the \(n\) fish are released.

  1. Find the value of \(n\).  (1 mark)

  2. Find the value of \(Q\) when \(t=6\).
  3. Give your answer correct to the nearest integer.  (1 mark)

  4.  i. Given that  \(\dfrac{dQ}{dt}=\dfrac{11}{10} Q\left(1-\dfrac{Q}{1000}\right)\),  express  \(\dfrac{d^2 Q}{d t^2}\)  in terms of \(Q\).  (1 mark)

  5. ii. Hence or otherwise, find the size of the fish population in pond 2 and the value of \(t\) when the rate of growth of the population is a maximum. Give your answer for \(t\) correct to the nearest year.  (2 marks)

  6. Sketch the graph of \(Q\) versus \(t\) on the set of axes below. Label any axis intercepts and any asymptotes with their equations.  (1 mark)
     

The farmer wishes to take 5.5% of the fish from pond 2 each year. The modified logistic differential equation that would model the fish population, \(Q\), in pond 2 after \(t\) years in this situation is

\(\dfrac{d Q}{d t}=\dfrac{11}{10}\, Q\left(1-\dfrac{Q}{1000}\right)-0.055Q\)

  1. Find the maximum number of fish that could be supported in pond 2 in this situation.  (1 mark)

Show Answers Only

a.    \(A=1, \quad B=\dfrac{1}{1000}\)

b.   \(\Rightarrow D=4\)

c.    \(n=\dfrac{1000}{1+9 e^0}=100\)

d.   \(Q=\dfrac{1000}{1+9 e^{-6.6}} \approx 988\)

e.i.  \(Q^{\prime}=\frac{121}{100}\, Q\left(1-\frac{Q}{1000}\right)^2-\frac{121}{100\ 000}\left(1-\frac{Q}{1000}\right)\)

e.ii. \(t=2\)

f.   


g.  \(Q=950\)

Show Worked Solution

a.    \(\dfrac{dP}{dt}=P\left(1-\dfrac{P}{1000}\right) \ \Rightarrow \ \dfrac{d t}{d P}=\dfrac{1}{P}\left(\dfrac{1}{1-\frac{P}{1000}}\right)\)

\(\text{Expand (by CAS):}\)

\(\dfrac{1}{P}\left(\dfrac{1}{1-\frac{P}{1000}}\right)=\dfrac{1}{P}-\dfrac{1}{(P-1000)}=\dfrac{1}{P}+\dfrac{1}{1000}\left(\dfrac{1}{1-\frac{P}{1000}}\right)\)

\(A=1, \quad B=\dfrac{1}{1000}\)
 

♦ Mean mark (a) 48%.

b.   \(\text{When}\ \ t=0, P=200 \text{ (given)}\)

\(\text{Solve}\ \ 200=\dfrac{1000}{1+D e^0}\ \ \text{for}\  t:\)

\(\Rightarrow D=4\ \ \text {(by CAS):}\)
 

c.    \(Q=\dfrac{1000}{1+9 e^{-1.1 t}}\)

\(\text{At}\ \ t=0, Q=n\):

\(n=\dfrac{1000}{1+9 e^0}=100\)
 

d.    \(\text{Find } Q \text{ when } t=6:\)

\(Q=\dfrac{1000}{1+9 e^{-6.6}} \approx 988\)
 

e.i.  \(\dfrac{d Q}{d t}=\dfrac{11}{10}\, Q\left(1-\dfrac{Q}{1000}\right)\)

\(\text{Using the product rule:}\)

\begin{aligned}
Q^{\prime \prime} & =\frac{11}{10}\left[Q^{\prime}\left(1-\frac{Q}{1000}\right)+Q\left(-\frac{1}{1000}\, Q^{\prime}\right)\right] \\
& =\frac{11}{10}\left[\frac{11}{10}\, Q\left(1-\frac{Q}{1000}\right)^2-\frac{Q}{1000}\left(\frac{11}{10}\, Q\left(1-\frac{Q}{1000}\right)\right)\right] \\
& =\frac{121}{100}\, Q\left(1-\frac{Q}{1000}\right)^2-\frac{121}{100\ 000}\left(1-\frac{Q}{1000}\right)
\end{aligned}

 

♦♦♦ Mean mark (e)(i) 21%.

e.ii. \(\text{Max } Q^{\prime} \Rightarrow Q^{\prime \prime}=0\)

\(\text{Solve } Q^{\prime \prime}=0 \ \ \text {(by CAS):}\)

\(\Rightarrow Q=500\)

\(\text{Solve } Q=500 \text{ for } t \text{ (by CAS):}\)

\(t=1.99 \ldots=2 \ \text{(nearest year)}\)
 

f.   


g.  \(\text{Solve for } Q \ \text{(by CAS):}\)

\(\dfrac{d Q}{d t}=\dfrac{11}{10}\, Q\left(1-\dfrac{Q}{1000}\right)-0.055\,Q=0\)

\(\Rightarrow Q=950\)

♦ Mean mark (g) 40%.

PHYSICS, M7 2023 VCE 13

A group of physics students undertake a Young's double-slit experiment using the apparatus shown in Figure 15. They use a green laser that produces light with a wavelength of 510 nm. The light is incident on two narrow slits, S\(_1\) and S\(_2\). The distance between the two slits is 100 \( \mu \)m.
 

An interference pattern is observed on a screen with points P\(_{0}\), P\(_{1}\) and P\(_2\) being the locations of adjacent bright bands, shown in Figure 15. Point  P\(_0\) is the central bright band.

  1. Calculate the path difference between S\(_{1}\)P\(_{2}\) and S\(_{2}\)P\(_{2}\). Give your answer in metres. Show your working.  (2 marks)
  1. The green laser is replaced by a red laser.
  2. Describe the effect of this change on the spacing between adjacent bright bands.  (1 mark)
  1. Explain how Young's double-slit experiment provides evidence for the wave-like nature of light and not the particle-like nature of light.  (3 marks)
Show Answers Only

a.    \(1.02 \times 10^{-6}\ \text{m}\)

b.    The spacing between adjacent bright bands will increase.

c.    Evidence for the wave-like nature of light:

→ Young’s double slit experiment was used to show that light can produce an interference pattern as it diffracts when it passes through the slits.

→ As this is a wave phenomenon, the experiment provided valuable evidence to support the wave-like nature of light.

→ Using the particle-like nature of light as a model, two bright bands would have been expected behind the two slits as the particles would have passed through the slits in a straight line which was not observed.

Show Worked Solution

a.    \(\text{Difference in distance}\ = 2 \lambda\)

\(\text{Path difference}\ =2 \times 510 \times 10^{-9} = 1.02 \times 10^{-6}\ \text{m}\).

♦ Mean mark (a) 45%.

b.    The spacing will increase.

→ \( x=\dfrac{m\lambda D}{d}\), where \(D\) is the length between the screen and the slits and \(x\) is the distance between adjacent bright bands.

→ Since \(x \propto \lambda\), as the wavelength of light increases from green to red, so will the spacing between adjacent bright bands.
 

c.    Evidence for the wave-like nature of light:

→ Young’s double slit experiment was used to show that light can produce an interference pattern as it diffracts when it passes through the slits.

→ As this is a wave phenomenon, the experiment provided valuable evidence to support the wave-like nature of light.

→ Using the particle-like nature of light as a model, two bright bands would have been expected behind the two slits as the particles would have passed through the slits in a straight line which was not observed.

PHYSICS, M7 2023 VCE 10

A proton in an accelerator beamline of proper length 4.80 km has a Lorentz factor, \(\gamma\), of 2.00.

  1. Calculate the speed of the proton relative to the beamline in terms of \(c\), the speed of light in a vacuum. Give your answer to three significant figures.  (3 marks)
  1. Calculate the length of the beamline in the reference frame of the proton.  (1 mark)
Show Answers Only

a.    \(0.866c\)

b.    \(2.4\ \text{km}\)

Show Worked Solution
a.   \(\dfrac{1}{\sqrt{1-\frac{v^2}{c^2}}}\) \(=2\)  
\(\sqrt{1-\dfrac{v^2}{c^2}}\) \(=\dfrac{1}{2}\)  
\(1-\dfrac{v^2}{c^2}\) \(=\dfrac{1}{4}\)  
\(\dfrac{v^2}{c^2}\) \(=\dfrac{3}{4}\)  
\(v^2\) \(=\dfrac{3}{4}c^2\)  
\(v\) \(=0.866c\)  

 

b.     \(l\) \(=l_0\sqrt{1-\dfrac{v^2}{c^2}}\)
    \(=4.8 \times \sqrt{1-\dfrac{(0.886c)^2}{c^2}}\)
    \(=4.8 \times 0.5\)
    \(=2.4\ \text{km}\)

PHYSICS, M5 2023 VCE 9

Giorgos is practising his tennis serve using a tennis ball of mass 56 g.

  1. Giorgos practises throwing the ball vertically upwards from point A to point B, as shown in Figure 10 . His daughter Eka, a physics student, models the throw, assuming that the ball is at the level of Giorgos's shoulder, point A, both when it leaves his hand and also when he catches it again. Point A is 1.8 m from the ground. The ball reaches a maximum height, point B, 1.8 m above Giorgos's shoulder.

  1. Show that the ball is in the air for 1.2 s from the time it leaves Giorgos's hand, which is level with his shoulder, until he catches it again at the same height.  (2 marks)
  1. Giorgos swings his racquet from point D through point C, which is horizontally behind him at shoulder height, as shown in Figure 11, to point B. Eka models this swing as circular motion of the racquet head. The centre of the racquet head moves with constant speed in a circular arc of radius 1.8 m from point C to point B.

  1. The racquet passes point C at the same time that the ball is released at point A and then the racquet hits the ball at point B.
  2. Calculate the speed of the racquet at point C.  (2 marks)
  1. The ball leaves Giorgos's racquet with an initial speed of 24 m s\(^{-1}\) in a horizontal direction, as shown in Figure 12. A tennis net is located 12 m in front of Giorgos and has a height of 0.90 m.
     

  1. How far above the net will the ball be when it passes above the net? Assume that there is no air resistance. Show your working.  (3 marks)

Show Answers Only

a.    \(1.2\ \text{s}\)

b.    \(4.7\ \text{ms}^{-1}\)

c.    \(1.475\ \text{m}\)

Show Worked Solution

a.    Time to for the ball to fall from max height back to shoulder level:

\(s\) \(=ut+\dfrac{1}{2}at^2\)  
\(1.8\) \(=0 \times t + \dfrac{1}{2} \times 9.8 \times t^2\)  
\(t\) \(=\sqrt{\dfrac{1.8}{4.9}}\)  
  \(=0.606\ \text{s}\)  

 

\(\text{Total time of flight}\ =0.606 \times 2=1.2\ \text{s}\)
 

b.    The racket travels a quarter of the circumference of the circle in 0.606 seconds.

\(v=\dfrac{d}{t}=\dfrac{2\pi \times 1.8 \times 0.25}{0.606}=4.7\ \text{ms}^{-1}\)
 

♦♦ Mean mark (b) 27%.

c.    Time for the ball to reach the net:

\(t=\dfrac{d}{v}=\dfrac{12}{24}=0.5\ \text{s}\)

Vertical displacement from max height in 0.5 seconds:

\(s=ut+\dfrac{1}{2}at^2=0 \times 0.5 + \dfrac{1}{2} \times 9.8 \times 0.5^2=1.225\ \text{m}\).

\(\text{Height (at net)}\ =3.6-1.225=2.375\ \text{m}\).

\(\therefore\ \text{Height (above net)}\ =2.375-0.9=1.475\ \text{m}\)

♦ Mean mark (c) 47%.

Calculus, SPEC2 2023 VCAA 3

The curve given by  \(y^2=x-1\), where  \(2 \leq x \leq 5\), is rotated about the \(x\)-axis to form a solid of revolution.

  1.  i. Write down the definite integral, in terms of \(x\), for the volume of this solid of revolution.  (1 mark)

  2. ii. Find the volume of the solid of revolution.  (1 mark)

  3.  i. Express the curved surface area of the solid in the form  \(\pi \displaystyle \int_a^b \sqrt{A x-B}\, d x\), where \(a, b, A, B\) are all positive integers.  (2 marks)

  4. ii. Hence or otherwise, find the curved surface area of the solid correct to three decimal places.  (1 mark)

The total surface area of the solid consists of the curved surface area plus the areas of the two circular discs at each end.

The 'efficiency ratio' of a body is defined as its total surface area divided by the enclosed volume.

  1. Find the efficiency ratio of the solid of revolution correct to two decimal places.  (2 marks)

  2. Another solid of revolution is formed by rotating the curve given by  \(y^2=x-1\) about the \(x\)-axis for  \(2 \leq x \leq k\), where \(k \in R\). This solid has a volume of \(24 \pi\).
  3. Find the efficiency ratio for this solid, giving your answer correct to two decimal places.  (3 marks)

Show Answers Only

a.i. \(\displaystyle V=\pi \int y^2 d x=\pi \int_2^5(x-1) dx\)
 

a.ii. \(\text{Evaluating integral (by calc):}\)

\(V=\dfrac{15 \pi}{2}\ \text{u}^3\)
 

b.i.  \(\displaystyle y=\sqrt{x-1} \ \Rightarrow \ \frac{d y}{d x}=\frac{1}{2} \cdot \frac{1}{\sqrt{x-1}} \ \Rightarrow \ \frac{d^2 y}{dx^2}=\frac{1}{4(x-1)}\)

\(\ \ \begin{aligned} \displaystyle \int_2^5 2 \pi \sqrt{x-1} \ \sqrt{1+\frac{1}{4(x-1)}} \ d x & =\int_2^5 2 \pi \sqrt{x-1+\frac{1}{4}}\, d x \\ & =\pi \int_2^5 \sqrt{4 x-3}\, d x\end{aligned}\)

 
b.ii. \(\text{Evaluate integral in b.i.}\)

\(\text {S.A.}=30.847\ \text{u}^2\)
 

c.    \(y=\sqrt{x-1}\)

\(\text{At}\ \ x=2\ \ \Rightarrow y=1\)

\(\text{At}\ \ x=5\ \ \Rightarrow y=2\)

\begin{aligned}
\text{Total S.A. } &=30.847+\pi(1)^2+\pi(2)^2 \\
&=46.5545\ \text{u}^2 \\
\end{aligned}

\(\text {Efficiency ratio}=\dfrac{46.5545}{\frac{15 \pi}{2}}=1.98\ \text{(2 d.p.)}\)
 

d.   \(V=\displaystyle \pi \int_2^k(x-1) d x=24 \pi\)

\(\text{Solve (by calc):}\)

\(\dfrac{\pi k(k-2)}{2}=24 \pi \ \Rightarrow \ k=8 \)

\(\text {S.A.}=\pi \displaystyle{\int_2^8} \sqrt{4 x-3}\, d x=75.9163 \ldots\)

\(\text{Total S.A.}=75.9163+\pi(1+7)=101.049 \)

\(\text{Efficiency ratio}=\dfrac{101.049}{24 \pi}= 1.34\ \text{(2 d.p.)}\)

Show Worked Solution

a.i. \(\displaystyle V=\pi \int y^2 d x=\pi \int_2^5(x-1) dx\)
 

a.ii. \(\text{Evaluating integral (by calc):}\)

\(V=\dfrac{15 \pi}{2}\ \text{u}^3\)
 

b.i.  \(\displaystyle y=\sqrt{x-1} \ \Rightarrow \ \frac{d y}{d x}=\frac{1}{2} \cdot \frac{1}{\sqrt{x-1}} \ \Rightarrow \ \frac{d^2 y}{dx^2}=\frac{1}{4(x-1)}\)

\(\ \ \begin{aligned} \displaystyle \int_2^5 2 \pi \sqrt{x-1} \ \sqrt{1+\frac{1}{4(x-1)}} \ d x & =\int_2^5 2 \pi \sqrt{x-1+\frac{1}{4}}\, d x \\ & =\pi \int_2^5 \sqrt{4 x-3}\, d x\end{aligned}\)

 
b.ii. \(\text{Evaluate integral in b.i.}\)

\(\text {S.A.}=30.847\ \text{u}^2\)
 

c.    \(y=\sqrt{x-1}\)

\(\text{At}\ \ x=2\ \ \Rightarrow y=1\)

\(\text{At}\ \ x=5\ \ \Rightarrow y=2\)

\begin{aligned}
\text{Total S.A. } &=30.847+\pi(1)^2+\pi(2)^2 \\
&=46.5545\ \text{u}^2 \\
\end{aligned}

\(\text {Efficiency ratio}=\dfrac{46.5545}{\frac{15 \pi}{2}}=1.98\ \text{(2 d.p.)}\)
 

d.   \(V=\displaystyle \pi \int_2^k(x-1) d x=24 \pi\)

\(\text{Solve (by calc):}\)

\(\dfrac{\pi k(k-2)}{2}=24 \pi \ \Rightarrow \ k=8 \)

\(\text {S.A.}=\pi \displaystyle{\int_2^8} \sqrt{4 x-3}\, d x=75.9163 \ldots\)

\(\text{Total S.A.}=75.9163+\pi(1+7)=101.049 \)

\(\text{Efficiency ratio}=\dfrac{101.049}{24 \pi}= 1.34\ \text{(2 d.p.)}\)

Complex Numbers, SPEC2 2023 VCAA 2

Let \(w=\text{cis}\left(\dfrac{2 \pi}{7}\right)\).

  1. Verify that \(w\) is a root of  \(z^7-1=0\).   (1 marks)
  1. List the other roots of  \(z^7-1=0\)  in polar form.   (1 mark)
  1. On the Argand diagram below, plot and label the points that represent all the roots of  \(z^7-1=0\).   (2 marks)

  1.  i. On the Argand diagram below, sketch the ray that originates at the real root of  \(z^7-1=0\)  and passes through the point represented by \( \text{cis}\left(\dfrac{2 \pi}{7} \right)\).   (1 mark)

     

  1. ii. Find the equation of this ray in the form \(\text{Arg}\left(z-z_0\right)=\theta\), where \(z_0 \in C\), and \(\theta\) is measured in radians in terms of \(\pi\).   (1 mark)

  2. Verify that the equation  \(z^7-1=0\)  can be expressed in the form 
  3.     \((z-1)\left(z^6+z^5+z^4+z^3+z^2+z+1\right)=0\).   (1 mark)
  4.   i. Express  \(\text{cis}\left(\dfrac{2 \pi}{7}\right)+\operatorname{cis}\left(\dfrac{12 \pi}{7}\right)\) in the form \(A \cos (B \pi)\), where \(A, B \in R^{+}\).   (1 mark)

  5. ii. Given that  \(w=\operatorname{cis}\left(\dfrac{2 \pi}{7}\right)\)  satisfies  \((z-1)\left(z^6+z^5+z^4+z^3+z^2+z+1\right)=0\),

use De Moivre's theorem to show that

      1. \(\cos \left(\dfrac{2 \pi}{7}\right)+\cos \left(\dfrac{4 \pi}{7}\right)+\cos \left(\dfrac{6 \pi}{7}\right)=-\dfrac{1}{2}\).   (2 marks)

Show Answers Only

a.   \(w^7-1 =\ \text{cis} \Big{(}\dfrac{2 \pi}{7} \Big{)}^7-1 = \text{cis} \Big{(}\dfrac{14 \pi}{7}\Big{)}-1 = 1-1=0\)

\(\therefore w\ \text{is a root of}\ \ z^7-1=0 \)
 

b.  \(\text{Roots of}\ \ z^7-1=0:\) 

\(\text{cis} (0), \text{cis} \Big{(}\dfrac{2 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{4 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{6 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{8 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{10 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{12 \pi}{7} \Big{)} \)

c.   
          

d.i.   
          

d.ii. \(\text{Base angle (isosceles Δ)}\ = \dfrac{1}{2} \times \Big{(}\pi-\dfrac{2\pi}{7}\Big{)} = \dfrac{5\pi}{14} \)

\(\arg(z-1)=\dfrac{9\pi}{14} \)
 

e.    \((z-1)(z^6+z^5+z^4+z^3+z^2+z+1) \)  
    \(=(z^7+z^6+z^5+z^4+z^3+z^2+z)-(z^6+z^5+z^4+z^3+z^2+z+1) \)  
    \(=z^7-1\)  

 

f.i.   \(\text{cis}\Big{(}\dfrac{2\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{12\pi}{7}\Big{)} = \text{cis}\Big{(}\dfrac{2\pi}{7}\Big{)} + \text{cis}\Big{(}-\dfrac{2\pi}{7}\Big{)} =2\cos\Big{(}\dfrac{2\pi}{7}\Big{)}  \)
 

f.ii.  \(z^6+z^5+z^4+z^3+z^2+z+1=0\ \ \text{(using part (e))} \)

\((z^6+z)+(z^4+z^3)+(z^5+z^2)=-1 \)

\(\Bigg{(}\text{cis}\Big{(}\dfrac{12\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{2\pi}{7}\Big{)}\Bigg{)} + \Bigg{(}\text{cis}\Big{(}\dfrac{8\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{6\pi}{7}\Big{)}\Bigg{)} + \Bigg{(}\text{cis}\Big{(}\dfrac{10\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{4\pi}{7}\Big{)}\Bigg{)}=-1\)

\(2\cos \Big{(}\dfrac{2 \pi}{7}\Big{)}+2\cos \Big{(}\dfrac{6 \pi}{7}\Big{)}+2\cos \Big{(}\dfrac{4 \pi}{7}\Big{)}=-1\)

\(\cos \Big{(}\dfrac{2 \pi}{7}\Big{)}+\cos \Big{(}\dfrac{4 \pi}{7}\Big{)}+\cos \Big{(}\dfrac{6 \pi}{7}\Big{)}=-\dfrac{1}{2}\)

Show Worked Solution

a.   \(w^7-1 =\ \text{cis} \Big{(}\dfrac{2 \pi}{7} \Big{)}^7-1 = \text{cis} \Big{(}\dfrac{14 \pi}{7}\Big{)}-1 = 1-1=0\)

\(\therefore w\ \text{is a root of}\ \ z^7-1=0 \)
 

b.  \(\text{Roots of}\ \ z^7-1=0:\) 

\(\text{cis} (0), \text{cis} \Big{(}\dfrac{2 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{4 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{6 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{8 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{10 \pi}{7} \Big{)}, \text{cis} \Big{(}\dfrac{12 \pi}{7} \Big{)} \)

c.   
          

d.i.   
          

d.ii. \(\text{Base angle (isosceles Δ)}\ = \dfrac{1}{2} \times \Big{(}\pi-\dfrac{2\pi}{7}\Big{)} = \dfrac{5\pi}{14} \)

\(\arg(z-1)=\dfrac{9\pi}{14} \)
 

e.    \((z-1)(z^6+z^5+z^4+z^3+z^2+z+1) \)  
    \(=(z^7+z^6+z^5+z^4+z^3+z^2+z)-(z^6+z^5+z^4+z^3+z^2+z+1) \)  
    \(=z^7-1\)  

 

f.i.   \(\text{cis}\Big{(}\dfrac{2\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{12\pi}{7}\Big{)} = \text{cis}\Big{(}\dfrac{2\pi}{7}\Big{)} + \text{cis}\Big{(}-\dfrac{2\pi}{7}\Big{)} =2\cos\Big{(}\dfrac{2\pi}{7}\Big{)}  \)
 

f.ii.  \(z^6+z^5+z^4+z^3+z^2+z+1=0\ \ \text{(using part (e))} \)

\((z^6+z)+(z^4+z^3)+(z^5+z^2)=-1 \)

\(\Bigg{(}\text{cis}\Big{(}\dfrac{12\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{2\pi}{7}\Big{)}\Bigg{)} + \Bigg{(}\text{cis}\Big{(}\dfrac{8\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{6\pi}{7}\Big{)}\Bigg{)} + \Bigg{(}\text{cis}\Big{(}\dfrac{10\pi}{7}\Big{)} + \text{cis}\Big{(}\dfrac{4\pi}{7}\Big{)}\Bigg{)}=-1\)

\(2\cos \Big{(}\dfrac{2 \pi}{7}\Big{)}+2\cos \Big{(}\dfrac{6 \pi}{7}\Big{)}+2\cos \Big{(}\dfrac{4 \pi}{7}\Big{)}=-1\)

\(\cos \Big{(}\dfrac{2 \pi}{7}\Big{)}+\cos \Big{(}\dfrac{4 \pi}{7}\Big{)}+\cos \Big{(}\dfrac{6 \pi}{7}\Big{)}=-\dfrac{1}{2}\)

Algebra, MET2 2023 VCAA 16 MC

Let \(f(x)=e^{x-1}\).

Given that the product function \(f(x)\times g(x)=e^{(x-1)^2}\), the rule for the function \(g\) is

  1. \(g(x)=e^{x-1}\)
  2. \(g(x)=e^{(x-2)(x-1)}\)
  3. \(g(x)=e^{(x+2)(x-1)}\)
  4. \(g(x)=e^{x(x-2)}\)
  5. \(g(x)=e^{x(x-3)}\)
Show Answers Only

\(B\)

Show Worked Solution
\(f(x)\times g(x)\) \(=e^{(x-1)^2}\)
\(e^{x-1}\times g(x)\) \(=e^{(x-1)^2}\)
\( g(x)\) \(=\dfrac{e^{(x-1)^2}}{e^{(x-1)}}\)
  \(=e^{(x-1)^2}\times e^{(x-1)^-1}\)
  \(=e^{x^2-3x+2}\)
  \(=e^{(x-2)(x-1)}\)

 
\(\Rightarrow B\)

Probability, MET2 2023 VCAA 4

A manufacturer produces tennis balls.

The diameter of the tennis balls is a normally distributed random variable \(D\), which has a mean of 6.7 cm and a standard deviation of 0.1 cm.

  1. Find  \(\Pr(D>6.8)\), correct to four decimal places.   (1 mark)
  2. Find the minimum diameter of a tennis ball that is larger than 90% of all tennis balls produced.
  3. Give your answer in centimetres, correct to two decimal places.   (1 mark)

Tennis balls are packed and sold in cylindrical containers. A tennis ball can fit through the opening at the top of the container if its diameter is smaller than 6.95 cm.

  1. Find the probability that a randomly selected tennis ball can fit through the opening at the top of the container.
  2. Give your answer correct to four decimal places.   (1 mark)
  3. In a random selection of 4 tennis balls, find the probability that at least 3 balls can fit through the opening at the top of the container.
  4. Give your answer correct to four decimal places.   (2 marks)

A tennis ball is classed as grade A if its diameter is between 6.54 cm and 6.86 cm, otherwise it is classed as grade B.

  1. Given that a tennis ball can fit through the opening at the top of the container, find the probability that it is classed as grade A.
  2. Give your answer correct to four decimal places.   (2 marks)
  3. The manufacturer would like to improve processes to ensure that more than 99% of all tennis balls produced are classed as grade A.
  4. Assuming that the mean diameter of the tennis balls remains the same, find the required standard deviation of the diameter, in centimetres, correct to two decimal places.   (2 marks)
  5. An inspector takes a random sample of 32 tennis balls from the manufacturer and determines a confidence interval for the population proportion of grade A balls produced.
  6. The confidence interval is (0.7382, 0.9493), correct to four decimal places.
  7. Find the level of confidence that the population proportion of grade A balls is within the interval, as a percentage correct to the nearest integer.   (2 marks)

A tennis coach uses both grade A and grade B balls. The serving speed, in metres per second, of a grade A ball is a continuous random variable, \(V\), with the probability density function
 

\(f(v) = \begin {cases}
\dfrac{1}{6\pi}\sin\Bigg(\sqrt{\dfrac{v-30}{3}}\Bigg)         &\ \ 30 \leq v \leq 3\pi^2+30 \\
0         &\ \ \text{elsewhere}
\end{cases}\)
 

  1. Find the probability that the serving speed of a grade A ball exceeds 50 metres per second.
  2. Give your answer correct to four decimal places.   (1 mark)
  3. Find the exact mean serving speed for grade A balls, in metres per second.   (1 mark)

The serving speed of a grade B ball is given by a continuous random variable, \(W\), with the probability density function \(g(w)\).

A transformation maps the graph of \(f\) to the graph of \(g\), where \(g(w)=af\Bigg(\dfrac{w}{b}\Bigg)\).

  1. If the mean serving speed for a grade B ball is \(2\pi^2+8\) metres per second, find the values of \(a\) and \(b\).   (2 marks)
Show Answers Only

a.    \(0.1587\)

b.    \(d\approx6.83\)

c.    \(0.9938\)

d.    \(0.9998\)

e.    \(0,8960\)

f.    \(0.06\)

g.    \(90\%\)

h.    \(0.1345\)

i.    \(3(\pi^2+4)\)

j.    \(a=\dfrac{3}{2}, b=\dfrac{2}{3}\)

Show Worked Solution

a.    \(\text{Normal distribution:}\to \mu=6.7, \sigma=0.1\)

\(D\sim N(6.7, 0.1^2)\)

\(\text{Using CAS: }[\text{normCdf}(6.8, \infty, 6.7, 0.1)]\)

\(\Pr(D>6.8)=0.15865…\approx0.1587\)
 

b.    \(\Pr(D<d)=0.90\)  

\(\Pr\Bigg(Z<\dfrac{d-6.7}{0.1}\Bigg)=0.90\)

\(\text{Using CAS: }[\text{invNorm}(0.9, 6.7, 0.1)]\)

\(\therefore\ d=6.828155…\approx6.83\ \text{cm}\)

 

c.   \(\text{Normal distribution:}\to \mu=6.7, \sigma=0.1\)

\(D\sim N(6.7, 0.1^2)\)

\(\text{Using CAS: }[\text{normCdf}(-\infty, 6.95, 6.7, 0.1)]\)

\(\Pr(D<6.95)=0.99379…\approx0.9938\)
 

d.    \(\text{Binomial:}\to  n=4, p=0.99379…\)

\(X=\text{number of balls}\)

\(X\sim \text{Bi}(4, 0.999379 …)\)

\(\text{Using CAS: }[\text{binomCdf}(4, 0.999379 …, 3, 4)]\)

\(\Pr(X\geq 3)=0.99977 …\approx 0.9998\)

 

e.    \(\Pr(\text{Grade A|Fits})\) \(=\Pr(6.54<D<6.86|D<6.95)\)
    \(=\dfrac{\Pr(6.54<D<6.86)}{\Pr(D<6.95)}\)
  \(\text{Using CAS: }\) \(=\Bigg[\dfrac{\text{normCdf}(6.54, 6.86, 6.7, 0.1)}{\text{normCdf}(-\infty, 6.95, 6.7, 0.1)}\Bigg]\)
    \(=\dfrac{0.89040…}{0.99977…}\)
    \(=0.895965…\approx 0.8960\)

  

f.    \(\text{Normally distributed → symmetrical}\)

\(\text{Pr ball diameter outside the 99% interval}=1-0.99=0.01\)

\(D\sim N(6.7, \mu^2)\)

\(\Pr(6.54<D<6.86)>0.99\)

\(\therefore\ \Pr(D<6.54)<\dfrac{1-0.99}{2}\)

\(\text{Find z score using CAS: }\Bigg[\text{invNorm}\Bigg(\dfrac{1-0.99}{2},0,1\Bigg)\Bigg]=-2.575829…\)

\(\text{Then solve:} \ \dfrac{6.54-6.7}{\sigma}<-2.575829…\)

\(\therefore\ \sigma<0.0621\approx 0.06\)


♦♦ Mean mark (f) 35%.
MARKER’S COMMENT: Incorrect responses included using \(\Pr(D<6.86)=0.99\). Many answers were accepted in the range \(0<\sigma\leq0.06\) as max value of SD was not asked for provided sufficient working was shown.

g.   \(\hat{p}=\dfrac{0.7382+0.9493}{2}=0.84375\)

\(\text{Solve the following simultaneous equations for }z, \hat{p}=0.84375\)

\(0.84375-z\sqrt{\dfrac{0.84375(1-0.84375)}{32}}\) \(=0.7382\)      \((1)\)
\(0.84375+z\sqrt{\dfrac{0.84375(1-0.84375)}{32}}\) \(=0.9493\) \((2)\)
\(\text{Equation}(2)-(1)\)    
\(2z\sqrt{\dfrac{0.84375(1-0.84375)}{32}}\) \(=0.2111\)  
\(z=\dfrac{0.10555}{\sqrt{\dfrac{0.84375(1-0.84375)}{32}}}\) \(=1.64443352\)  

 
\(\text{Alternatively using CAS: Solve the following simultaneous equations for }z, \hat{p}\)

\(\hat{p}-z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{32}}\) \(=0.7382\)
\(\text{and}\)  
\(\hat{p}+z\sqrt{\dfrac{\hat{p}(1-\hat{p})}{32}}\) \(=0.9493\)

\(\rightarrow \ z=1.64444…\ \text{and}\ \hat{p}=0.84375\)
 

\(\therefore\ \text{Using CAS: normCdf}(-1.64443352, 1.64443352, 0 , 1)\)

\(\text{Level of Confidence} =0.89999133…=90\%\)


♦♦ Mean mark (g) 25%.
MARKER’S COMMENT: \(\hat{p}\) was often calculated incorrectly with \(\hat{p}=0.8904\) frequently seen.

h.   \(\text{Using CAS: Evaluate }\to \displaystyle \int_{50}^{3\pi^2+30} \frac{1}{6\pi}\sin\Bigg(\sqrt{\dfrac{v-30}{3}}\Bigg)\,dx=0.1345163712\approx 0.1345\)

i.   \(\text{Using CAS: Evaluate }\to \displaystyle \int_{30}^{3\pi^2+30} v.\dfrac{1}{6\pi}\sin\Bigg(\sqrt{\dfrac{v-30}{3}}\Bigg)\,dx=3(\pi^2+4)=3\pi^2+12\)

j.  \(\text{When the function is dilated in both directions, }a\times b=1\)
 

\(\text{Method 1 : Simultaneous equations}\)

\(g(w) = \begin {cases}
\dfrac{b}{6\pi}\sin\Bigg(\sqrt{\dfrac{\dfrac{w}{b}-30}{3}}\Bigg)         &\ \ 30b \leq v \leq b(3\pi^2+30) \\
\\ 0         &\ \ \text{elsewhere}
\end{cases}\)

\(\text{Using CAS: Define }g(w)\ \text{and solve the simultaneous equations below for }a, b.\)
 

\(\displaystyle \int_{30.b}^{b.(3\pi^2+30)} g(w)\,dw=1\)

\(\displaystyle \int_{30.b}^{b.(3\pi^2+30)} w.g(w)\,dw=2\pi^2+8\)

\(\therefore\ b=\dfrac{2}{3}\ \text{and}\ a=\dfrac{3}{2}\)

 
\(\text{Method 2 : Transform the mean}\)

 

\(\text{Area}\) \(=1\)
\(\therefore\ a\) \(=\dfrac{1}{b}\)
\(\to\ b\) \(=\dfrac{E(W)}{E(V)}\)
  \(=\dfrac{2\pi^2+8}{3\pi^2+12}\)
  \(=\dfrac{2(\pi^2+4)}{3(\pi^2+4)}\)
  \(=\dfrac{2}{3}\)
 
\(\therefore\ a\) \(=\dfrac{3}{2}\)

♦♦♦ Mean mark (j) 10%.

Calculus, SPEC2 2023 VCAA 1

Viewed from above, a scenic walking track from point \(O\) to point \(D\) is shown below. Its shape is given by

\(f(x)= \begin{cases}-x(x+a)^2, & 0 \leq x \leq 1 \\ e^{x-1}-x+b, & 1<x \leq 2 .\end{cases}\)

The minimum turning point of section \(O A B C\) occurs at point \(A\). Point \(B\) is a point of inflection and the curves meet at point \(C(1,0)\). Distances are measured in kilometres.
 

  1. Show that  \(a=-1\)  and  \(b=0\).  (1 mark)

  2. Verify that the two curves meet smoothly at point \(C\).  (2 marks)

  3.  i. Find the coordinates of point \(A\).  (1 mark)

  4. ii. Find the coordinates of point \(B\).  (1 mark)

The return track from point \(D\) to point \(O\) follows an elliptical path given by

\(x=2 \cos (t)+2, y=(e-2) \sin (t)\), where \(t \in\left[\dfrac{\pi}{2}, \pi\right]\).

  1. Find the Cartesian equation of the elliptical path.  (2 marks)

  2. Sketch the elliptical path from \(D\) to \(O\) on the diagram above.  (1 mark)

  3.  i. Write down a definite integral in terms of \(t\) that gives the length of the elliptical path from \(D\) to \(O\).  (1 mark)

  4. ii. Find the length of the elliptical path from \(D\) to \(O\).
  5.     Give your answer in kilometres correct to three decimal places.   (1 mark)

Show Answers Only

a.    \(\text{At}\ C(1,0): \)

\(\underset{x \to 1^{-}}{\lim} f_1(x) = \underset{x \to 1^{-}}{\lim}-x(x+a)^2 = -1(1+a)^2 = 0 \)

    \(a=-1\)

\(\underset{x \to 1^{+}}{\lim} f_2(x)\ \underset{x \to 1^{+}}{\lim} (e^{x-1}-x+b) = e^0-1+b=b = 0 \)
 

b.    \(\text{Find derivative functions (by calc)}:\)

\(\dfrac{d}{dx} \big{(}-x(x-1)^2\big{)} = -(x-1)(3x-1)\ \ \Rightarrow \ \ \text{At}\ \ x=1, f_1^{′}(x)=0 \)

\(\dfrac{d}{dx} (e^{x-1}-x+b) = e^{x-1}-1\ \ \Rightarrow \ \ \text{At}\ \ x=1, f_2^{′}(x)=0 \)

\(\Rightarrow\ \text{Since gradients are equal, curves meet smoothly.}\)
 

c.i.  \(A\Big{(} \frac{1}{3}, -\frac{4}{27} \Big{)} \)
 

c.ii.  \(B\Big{(} \frac{2}{3}, -\frac{2}{27} \Big{)} \)
 

d.   \(\Big{(}\dfrac{x-2}{2}\Big{)}^2 + \Big{(}\dfrac{y}{e-2}\Big{)}^2=1\)
 

e.    
          
 

f.i.    \(\displaystyle{\int_{\frac{\pi}{2}}^{\pi}} \sqrt{4\,\sin^2(t)+(e-2)^2\cos^2(t)}\,dt\)
 

f.ii.  \(\text{Length}\ = 2.255\)

Show Worked Solution

a.    \(\text{At}\ C(1,0): \)

\(\underset{x \to 1^{-}}{\lim} f_1(x) = \underset{x \to 1^{-}}{\lim}-x(x+a)^2 = -1(1+a)^2 = 0 \)

    \(a=-1\)

\(\underset{x \to 1^{+}}{\lim} f_2(x)\ \underset{x \to 1^{+}}{\lim} (e^{x-1}-x+b) = e^0-1+b=b = 0 \)
 

b.    \(\text{Find derivative functions (by calc)}:\)

\(\dfrac{d}{dx} \big{(}-x(x-1)^2\big{)} = -(x-1)(3x-1)\ \ \Rightarrow \ \ \text{At}\ \ x=1, f_1^{′}(x)=0 \)

\(\dfrac{d}{dx} (e^{x-1}-x+b) = e^{x-1}-1\ \ \Rightarrow \ \ \text{At}\ \ x=1, f_2^{′}(x)=0 \)

\(\Rightarrow\ \text{Since gradients are equal, curves meet smoothly.}\)
 

c.i.  \(\dfrac{d}{dx} \big{(}-x(x-1)^2\big{)} = -(x-1)(3x-1)\ \Rightarrow \ \text{SP when}\ x=\dfrac{1}{3} \)

\(A\Big{(} \frac{1}{3}, -\frac{4}{27} \Big{)} \)
 

c.ii.  \(\text{POI at}\ B: \ f_1^{″}(x)=-6x+4=0\ \ \Rightarrow x=\dfrac{2}{3} \)

\(B\Big{(} \frac{2}{3}, -\frac{2}{27} \Big{)} \)
 

d.   \(\dfrac{x-2}{2}=\cos(t), \ \ \dfrac{y}{e-2}=\sin(t) \)

\(\Big{(}\dfrac{x-2}{2}\Big{)}^2 + \Big{(}\dfrac{y}{e-2}\Big{)}^2=1\)
 

e.    
          
 

f.i.    \(\dfrac{dx}{dt} = -2\sin(t),\ \ \dfrac{dy}{dt} = (e-2)\cos(t) \)

\(\displaystyle{\int_{\frac{\pi}{2}}^{\pi}} \sqrt{4\,\sin^2(t)+(e-2)^2\cos^2(t)}\,dt\)
 

f.ii.   \(\text{Evaluate the integral in part f.i.}\)

\(\Rightarrow \text{Length}\ = 2.255\)

PHYSICS, M6 2023 VCE 7*

Two high-voltage transmission lines span a distance of 260 km between Power Plant A and Town B, as shown in Figure 7. Power Plant A provides 350 MW of power. The potential difference at Power Plant A is 500 kV. The current in the transmission lines has a value of 700 A and the power loss in the transmission lines is 20 MW.
 

   

  1. Show, using calculations, that the total resistance of the two transmission lines is 41 \(\Omega\).  (2 marks)
  1. Town B needs a minimum of 480 kV.
  2. Determine whether 480 kV will be available to Town B. Show your working.  (3 marks)
  1. Explain what would happen if the electricity between Power Plant A and Town B were to be transmitted at 50 kV instead of 500 kV. Assume that the resistance of the transmission lines is still 41 \(\Omega\).  (2 marks)
Show Answers Only

a.    \(41\ \Omega\)

b.    The power generated in the power plant is 500 kV.

\(V_{\text{drop}}=IR=700 \times 41 =28.7\ \text{kV} \)

\(V_{\text{remaining}} = 500-28.7 = 471.3\ \text{kV} \)

\(\therefore\) 480V will not be available to Town B.
 

c.    If the voltage was decreased by a factor of 10 to 50 kV:

→ The current through the power lines would increase by a factor of 10.

→ Power remains constant \((P=VI)\)

→ As \(P_{\text{loss}}=I^2r\), this would dramatically increase the power loss and significantly less power would be transmitted to the town.

Show Worked Solution
a.     \(P_{\text{loss}}\) \(=I^2R\)
  \(R\) \(=\dfrac{P_{\text{loss}}}{I^2}\)
    \(=\dfrac{20 \times 10^6}{700^2}\)
    \(=41\ \Omega\)

 

b.    The power generated in the power plant is 500 kV.

\(V_{\text{drop}}=IR=700 \times 41 =28.7\ \text{kV} \)

\(V_{\text{remaining}} = 500-28.7 = 471.3\ \text{kV} \)

\(\therefore\) 480V will not be available to Town B.

 

c.    If the voltage was decreased by a factor of 10 to 50 kV:

→ Power remains constant \((P=VI)\)

→ The current through the power lines would increase by a factor of 10.

→ As \(P_{\text{loss}}=I^2r\), this would dramatically increase the power loss and significantly less power would be transmitted to the town.

PHYSICS, M6 2023 VCE 5

Figure 4a shows a single square loop of conducting wire placed just outside a constant uniform magnetic field, \(B\). The length of each side of the loop is 0.040 m. The magnetic field has a magnitude of 0.30 T and is directed out of the page.

Over a time period of 0.50 s, the loop is moved at a constant speed, \(v\), from completely outside the magnetic field, Figure 4a, to completely inside the magnetic field, Figure 4b.
 

  1. Calculate the average EMF produced in the loop as it moves from the position just outside the region of the field, Figure \(4 \mathrm{a}\), to the position completely within the area of the magnetic field, Figure 4b.
  2. Show your working.  (2 marks)
  1. On the small square loop in Figure 5, show the direction of the induced current as the loop moves into the area of the magnetic field.  (1 mark)
     

Show Answers Only

a.    \(9.6 \times 10^{-4}\ \text{V}\)

b.    
       

Show Worked Solution
a.     \(\varepsilon\) \(=\dfrac{\Delta \phi}{\Delta t}\)
    \(=\dfrac{\Delta B A}{\Delta t}\)
    \(=\dfrac{0.3 \times 0.04^2}{0.5}\)
    \(=9.6 \times 10^{-4}\ \text{V}\)

 

b.  
         

♦ Mean mark (b) 45%.

Calculus, MET2 2023 VCAA 5

Let \(f:R \to R, f(x)=e^x+e^{-x}\) and \(g:R \to R, g(x)=\dfrac{1}{2}f(2-x)\).

  1. Complete a possible sequence of transformations to map \(f\) to \(g\).   (2 marks)
    •    Dilation of factor \(\dfrac{1}{2}\) from the \(x\) axis.

Two functions \(g_1\) and \(g_2\) are created, both with the same rule as \(g\) but with distinct domains, such that \(g_1\) is strictly increasing and \(g_2\) is strictly decreasing.

  1. Give the domain and range for the inverse of \(g_1\).   (2 marks)

Shown below is the graph of \(g\), the inverse of \(g_1\) and \(g_2\), and the line \(y=x\).
 

The intersection points between the graphs of \(y=x, y=g(x)\) and the inverses of \(g_1\) and \(g_2\), are labelled \(P\) and \(Q\).

    1. Find the coordinates of \(P\) and \(Q\), correct to two decimal places.   (1 mark)
    1. Find the area of the region bound by the graphs of \(g\), the inverse of \(g_1\) and the inverse of \(g_2\).
    2. Give your answer correct to two decimal places.   (2 marks)

Let \(h:R\to R, h(x)=\dfrac{1}{k}f(k-x)\), where \(k\in (o, \infty)\).

  1. The turning point of \(h\) always lies on the graph of the function \(y=2x^n\), where \(n\) is an integer.
  2. Find the value of \(n\).  (1 mark)

Let \(h_1:[k, \infty)\to R, h_1(x)=h(x)\).

The rule for the inverse of \(h_1\) is \(y=\log_{e}\Bigg(\dfrac{1}{k}x+\dfrac{1}{2}\sqrt{k^2x^2-4}\Bigg)+k\)

  1. What is the smallest value of \(k\) such that \(h\) will intersect with the inverse of \(h_1\)?\
  2. Give your answer correct to two decimal places.   (1 mark)

It is possible for the graphs of \(h\) and the inverse of \(h_1\) to intersect twice. This occurs when \(k=5\).

  1. Find the area of the region bound by the graphs of \(h\) and the inverse of \(h_1\), where \(k=5\).
  2. Give your answer correct to two decimal places.   (2 marks)

Show Answers Only

a.    \(\text{Reflect in the }y\text{-axis}\)

\(\text{Then translate 2 units to the right}\)

\(\text{OR}\)

\(\text{Translate 2 units to the left}\)

\(\text{Then reflect in the }y\text{-axis}\)

\(\text{OR}\)

\(\text{Translate 2 units to the right only }(f \ \text{even function})\)

b.    \(\text{See worked solution}\)

c.i.  \(P(1.27, 1.27\), Q(4.09, 4.09)\)

c.ii. \(2\displaystyle \int_{1.27…}^{4.09…} (x-g(x))\,dx=5.56\)

d.    \(n=-1\)

e. \(k\approx 1.27\)

f. \(A=43.91\)

Show Worked Solution

a.    \(\text{Reflect in the }y\text{-axis}\)

\(\text{Then translate 2 units to the right}\)

\(\text{OR}\)

\(\text{Translate 2 units to the left}\)

\(\text{Then reflect in the }y\text{-axis}\)

\(\text{OR}\)

\(\text{Translate 2 units to the right only }(f \ \text{even function})\)

b.    \(\text{Some options include:}\)

•  \(\text{Domain}\to [1, \infty)\ \ \ \text{Range}\to [2, \infty)\)
\(\text{or}\)
•  \(\text{Domain}\to (1, \infty)\ \ \ \text{Range}\to (2, \infty)\)
 

\(\text{If functions split at turning point } (2, 1)\ \text{then possible options:}\)

• \(\text{Domain}\ g_1\to [2, \infty)\ \ \ \text{Range}\to [1, \infty)\)
   \(\text{and }\)
  \(\text{Domain}\ g_1^{-1}\to [1, \infty)\ \ \ \text{Range}\to [2, \infty)\)

\(\text{or}\)

•  \(\text{Domain}\ g_1\to (2, \infty)\ \ \ \text{Range}\to (1, \infty)\)
   \(\text{and }\)
   \(\text{Domain}\ g_1^{-1}\to (1, \infty)\ \ \ \text{Range}\to (2, \infty)\)


♦ Mean mark 50%.
MARKER’S COMMENT: Note: maximal domain not requested so there were many correct answers. Many students seemed to be confused by notation. Often domain and range reversed.

c.i.  \(\text{By CAS:}\  P(1.27, 1.27), Q(4.09, 4.09)\)

c.ii.   \(\text{Area}\) \(=2\displaystyle \int_{1.27…}^{4.09…} (x-g(x))\,dx\)
    \(=2\displaystyle \int_{1.27…}^{4.09…} \Bigg(x-\dfrac{1}{2}\Big(e^{2-x}+e^{x-2}\Big)\Bigg)\,dx\)
    \(\approx 5.56\)

♦♦ Mean mark (c)(ii) 30%.
MARKER’S COMMENT: Some students set up the integral but did not provide an answer.
Others did not double the integral. Some incorrectly used \(\displaystyle \int_{1.27…}^{4.09…}(g_1^-1-g(x))\,dx\).
Others had correct answer but incorrect integral.

d.    \((0, 2)\ \text{turning pt of }f(x)\)

\(\text{and }h(x) \text{is the transformation from }f(x)\)

\(\therefore \Bigg(k, \dfrac{2}{k}\Bigg)\ \text{turning tp of }h(x)\)

\(\text{so the coordinates have the relationship}\)

\(y=\dfrac{2}{x}=2x^{-1}\)

\(\therefore\ n=-1\)


♦♦♦ Mean mark (d) 10%.
MARKER’S COMMENT: Question not well done. \(n=1\) a common error.

e.    \(\text{Smallest }k\ \text{is when inverse of the curves }h_1\ \text{and}\ h\)

\(\text{touch with the line }y=x.\)

\(\therefore\ h(x)\) \(=\dfrac{1}{k}\Big(e^{k-x}+e^{x-k}\Big)\)
  \(=x\)

 

\(\text{and}\ h'(x)\) \(=\dfrac{1}{k}\Big(-e^{k-x}+e^{x-k}\Big)\)
  \(=1\)

\(\text{at point of intersection, where }k>0.\)

\(\text{Using CAS graph both functions to solve or solve as simultaneous equations.}\)

\(\rightarrow\ k\approx 1.2687\approx 1.27\)


♦♦♦♦ Mean mark (e) 0%.
MARKER’S COMMENT: Question poorly done with many students not attempting.

f.    \(\text{When}\ k=5\)

\(h_1^{-1}(x)=\log_e\Bigg(\dfrac{5}{2}x+\dfrac{1}{2}\sqrt{25x^2-4}\Bigg)+5\)

\(\text{and }h(x)=\dfrac{1}{5}\Big(e^{5-x}+e^{x-5}\Big)\)

\(\text{Using CAS: values of }x\ \text{at of intersection of fns are}\)

\(x\approx 1.45091…\ \text{and}\ 8.78157…\)

\(\text{Area}\) \(=\displaystyle \int_{1.45091…}^{8.78157…}h_1^{-1}(x)- h(x)\,dx\)
  \(=\displaystyle \int_{1.45091…}^{8.78157…} \log_e\Bigg(\dfrac{5}{2}x+\dfrac{1}{2}\sqrt{25x^2-4}\Bigg)+5-\dfrac{1}{5}\Big(e^{5-x}+e^{x-5}\Big)\,dx\)
  \(=43.91\) 

♦♦♦ Mean mark (f) 15%.
MARKER’S COMMENT: Errors were made with terminals. Common error using \(x=2.468\), which was the \(x\) value of the pt of intersection of \(h\) and \(y=x\).

PHYSICS, M6 2023 VCE 4*

A transformer is used to provide a low-voltage supply for six outdoor garden globes. The circuit is shown in Figure 3. Assume there is no power loss in the connecting wires. 
 

The input of the transformer is connected to a power supply that provides an AC voltage of 240 V. The globes in the circuit are designed to operate with an AC voltage of 12 V. Each globe is designed to operate with a power of 20 W.

  1. Assuming that the transformer is ideal, calculate the ratio of primary turns to secondary turns of the transformer.  (1 mark)

The globes are turned on.

  1. Calculate the current in the primary coil of the transformer.  (2 marks)
  1. Explain why the input current to the primary coil of the transformer must be AC rather than constant DC for the globes to shine.  (2 marks)

Show Answers Only

a.    \(20:1\)

b.    \(0.5\ \text{A}\)

c.    → A change in flux from the primary coil induces an EMF in the secondary coil.

→ Only an AC current will provide the change in flux required for the transformer to work. 

→ DC currents provide a constant flux and therefore will not work on a transformer.

Show Worked Solution

a.    \(\dfrac{N_p}{N_s}=\dfrac{240}{12}=20:1\)
 

b.    The total power drawn on the secondary side \(=6 \times 20=120\ \text{W}\)

Total power on the primary side is \(120\ \text{W}\).

\(I=\dfrac{P}{V}=\dfrac{120}{240}=0.5\ \text{A}\)

♦ Mean mark (b) 43%.

c.    → A change in flux from the primary coil induces an EMF in the secondary coil.

→ Only an AC current will provide the change in flux required for the transformer to work. 

→ DC currents provide a constant flux and therefore will not work on a transformer.

♦ Mean mark (c) 42%.

PHYSICS, M7 2023 VCE 19*

The diagram below shows the spectrum of light emitted by a hydrogen vapour lamp. The spectral line indicated by the arrow on the diagram is in the visible region of the spectrum.

   

Calculate the frequency of the light corresponding to the spectral line indicated by the arrow.   (2 marks)

Show Answers Only

\(B\)

Show Worked Solution

\(\lambda = 655\ \text{nm}\ = 655 \times 10^{-9}\ \text{m}\)

\(f=\dfrac{c}{\lambda}=\dfrac{3 \times 10^8}{655 \times 10^{-9}} \approx 4.6 \times 10^{14}\ \text{Hz}\)

PHYSICS, M7 2023 VCE 18 MC

Which one of the following statements best describes the type of light produced from different types of light sources?

  1. Light from a laser is coherent and has a very narrow range of wavelengths.
  2. Light from an incandescent lamp is coherent and has a range of wavelengths.
  3. Light from an incandescent lamp is incoherent and has a very narrow range of wavelengths.
  4. Light from a single-colour light-emitting diode (LED) is coherent and contains a very wide range of wavelengths.
Show Answers Only

\(A\)

Show Worked Solution

→ Laser light is coherent and usually monochromatic, meaning it produces a specific wavelength of light.

→ Incandescent light is incoherent but displays a range of wavelengths, therefore it is not \(B\) or \(C\).

→ LED’s contain a narrow range of wavelengths, therefore it is not \(D\).

\(\Rightarrow A\)

PHYSICS, M8 2023 VCE 17 MC

Which one of the following statements best explains why it is possible to compare X-ray and electron diffraction patterns?

  1. X-rays can exhibit particle-like properties.
  2. Electrons can exhibit wave-like properties.
  3. Electrons are a form of high-energy X-rays.
  4. Both electrons and X-rays can ionise matter.
Show Answers Only

\(B\)

Show Worked Solution

→ de Broglie developed an equation linking an objects momentum to its wavelength. 

→ At high speeds, the wavelength of electrons is similar to that of X-rays and so they can be compared.

\(\Rightarrow B\)