ENGAA 2016 SPECIMEN PAPER SECTION 1: Questions & Worked Solutions

Question 1

1 mark

A square piece of metal has a semicircular piece cut out of it as shown. The area of the remaining metal is 100 cm².

[diagram not to scale]

Which one of the following is a correct expression for the length of the side of the square in centimetres?

A. $20 \sqrt{\frac{2}{8-\pi}}$
B. $10 \sqrt{\frac{2}{4-\pi}}$
C. $20 \sqrt{\frac{2}{8+\pi}}$
D. $10 \sqrt{\frac{1}{8-\pi}}$
E. $20 \sqrt{\frac{1}{4-\pi}}$

Answer: A

Worked solution

Let the side length of the square be

s

. The radius of the semicircle is

r = s/2

. The area of the remaining metal is the area of the square minus the area of the semicircle.

\text{Area} = s^2 - \frac{1}{2}\pi r^2 \ 100 = s^2 - \frac{1}{2}\pi \left(\frac{s}{2}\right)^2 \ 100 = s^2 - \frac{\pi s^2}{8} \ 100 = s^2 \left(1 - \frac{\pi}{8}\right) \ 100 = s^2 \left(\frac{8 - \pi}{8}\right) \ s^2 = \frac{800}{8 - \pi} \ s = \sqrt{\frac{800}{8 - \pi}} \ s = \sqrt{\frac{400 \times 2}{8 - \pi}} \ s = 20\sqrt{\frac{2}{8 - \pi}}

Question 2

1 mark

Shortly after opening her parachute, a free-fall parachutist of mass 60 kg (including equipment) experiences the forces shown in the diagram.

drag (air resistance) = 900 N
weight = 600 N

[diagram not to scale]

Which line in the table gives the size and direction of the acceleration of the parachutist at this instant?

A.size of acceleration/ms⁻²: 5.0, direction of acceleration: downwards
B.size of acceleration/ms⁻²: 10.0, direction of acceleration: downwards
C.size of acceleration/ms⁻²: 5.0, direction of acceleration: upwards
D.size of acceleration/ms⁻²: 10.0, direction of acceleration: upwards
E.size of acceleration/ms⁻²: 0.0, direction of acceleration: —

Answer: C

Worked solution

The resultant force (

F_{\text{net}}

) is the difference between the upward drag and the downward weight.

F_{\text{net}} = 900 \, \text{N} - 600 \, \text{N} = 300 \, \text{N}

The result is positive, so the net force is directed upwards. According to Newton's second law,

F = ma

. The mass (

m

) is

60 \, \text{kg}

.

300 = 60 \times a \ a = \frac{300}{60} \ a = 5 \, \text{ms}^{-2}

The acceleration is in the same direction as the resultant force. Therefore, the acceleration is

5.0 \, \text{ms}^{-2}

upwards.

Question 3

1 mark

In a right-angled triangle PQR the hypotenuse is the side PR.

The length of side PQ is 20 cm and the ratio RQ:PQ is 1:2

What is the length of the perpendicular from the hypotenuse to the point Q?

A. $8\sqrt{5}$ cm
B. $10\sqrt{2}$ cm
C. $2\sqrt{5}$ cm
D. $5\sqrt{2}$ cm
E. $4\sqrt{5}$ cm

Answer: E

Worked solution

Given a right-angled triangle PQR with hypotenuse PR, we have:

-

PQ = 20 \text{ cm}

-

RQ : PQ = 1 : 2 \implies RQ = \frac{1}{2} \times 20 = 10 \text{ cm}

First, we find the length of the hypotenuse

PR

using Pythagoras' theorem:

PR^2 = PQ^2 + RQ^2

PR^2 = 20^2 + 10^2 = 400 + 100 = 500

PR = \sqrt{500} = 10\sqrt{5} \text{ cm}

Let

h

be the perpendicular distance from point

Q

to the hypotenuse. The area of the triangle can be calculated in two ways. We equate them to find

h

:

\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}

\frac{1}{2} \times RQ \times PQ = \frac{1}{2} \times PR \times h

\frac{1}{2} \times 10 \times 20 = \frac{1}{2} \times 10\sqrt{5} \times h

100 = 5\sqrt{5} \cdot h

h = \frac{100}{5\sqrt{5}} = \frac{20}{\sqrt{5}}

h = \frac{20\sqrt{5}}{5} = 4\sqrt{5} \text{ cm}

Question 4

1 mark

The displacement/time graph shown represents a wave of wavelength 1.5 cm.

What is the speed of the wave?

A. $0.33 \text{ cm s}^{-1}$
B. $0.67 \text{ cm s}^{-1}$
C. $0.75 \text{ cm s}^{-1}$
D. $1.33 \text{ cm s}^{-1}$
E. $1.5 \text{ cm s}^{-1}$
F. $3.0 \text{ cm s}^{-1}$

Answer: C

Worked solution

The relationship between wave speed (

c

), frequency (

f

), and wavelength (

\lambda

) is given by the formula:

c = f\lambda

From the displacement-time graph, the period (

T

), the time for one full wave cycle, is 2 seconds. The frequency is the reciprocal of the period:

f = \frac{1}{T} = \frac{1}{2} \, \text{Hz}

The wavelength is given as

\lambda = 1.5 \, \text{cm}

. Substituting these values into the wave speed equation:

c = \frac{1}{2} \times 1.5 = 0.75 \, \text{cm s}^{-1}

Question 5

1 mark

A cube has sides of unit length. What is the length of a line joining a vertex to the midpoint of one of the opposite faces (the dashed line in the diagram below)?

[diagram not to scale]

A. $\frac{\sqrt{3}}{2}$
B. $\sqrt{2}$
C. $\frac{\sqrt{5}}{2}$
D. $\sqrt{3}$
E. $\sqrt{5}$

Answer: A

Worked solution

The cube has sides of unit length. We can solve this by forming a right-angled triangle in 3D.

- One leg is a vertical edge of the cube, with length 1.

- The other leg is the line from a vertex on the base to the center of the base.

First, find the length of the diagonal of the square base:

\text{Diagonal}_{\text{base}} = \sqrt{1^2 + 1^2} = \sqrt{2}

The distance from a vertex to the center of the base is half of the diagonal’s length:

\text{Leg}_2 = \frac{\sqrt{2}}{2} = \frac{1}{\sqrt{2}}

Let the required length be

x

. Using Pythagoras’ Theorem on our 3D triangle:

x^2 = 1^2 + \left(\frac{1}{\sqrt{2}}\right)^2 = 1 + \frac{1}{2} = \frac{3}{2}

x = \sqrt{\frac{3}{2}}

Question 6

1 mark

A point mass travelling at a constant speed has a momentum of 30 N s and a kinetic energy of 150 J.

What is the mass of the object?

A.2 kg
B.3 kg
C.5 kg
D.6 kg
E.10 kg
F.15 kg

Answer: B

Worked solution

The momentum (

p

) and kinetic energy (

E_k

) are given by the equations:

p = mv = 30

E_k = \frac{1}{2} mv^2 = 150

From the kinetic energy equation, we can write:

mv^2 = 2 \times 150 = 300

Now, we can find the velocity (

v

) by dividing:

\frac{mv^2}{mv} = \frac{300}{30}

v = 10 \text{ m/s}

Substitute the value of

v

back into the momentum equation to find the mass (

m

):

m \times 10 = 30

m = \frac{30}{10}

m = 3 \text{ kg}

Question 7

1 mark

If you look at a clock and the time is 9.45, what is the angle between the hour and the minute hands?

A. $0^\circ$
B. $7.5^\circ$
C. $15^\circ$
D. $22.5^\circ$
E. $30^\circ$

Answer: D

Worked solution

At the time 9:45, the minute hand points exactly at the '9' on the clock face. The hour hand moves continuously. In 12 hours, it moves

360^\circ

. A time of 45 minutes is equivalent to

\frac{45}{60} = \frac{3}{4}

of an hour. Therefore, at 9:45, the hour hand has moved

\frac{3}{4}

of the way from the '9' towards the '10'. The angle between any two consecutive hour marks is

\frac{360^\circ}{12} = 30^\circ

. The angle between the minute hand (at '9') and the hour hand is the angle the hour hand has moved past the '9'.

\text{Angle} = \frac{3}{4} \times 30^\circ = \frac{90^\circ}{4} = 22.5^\circ

Question 8

1 mark

Which of the following is a correct unit of potential difference (voltage)?

A.amp per ohm
B.coulomb per joule
C.joule per second
D.newton per coulomb
E.watt per amp

Answer: E

Worked solution

We check the units derived from fundamental electrical equations for each option to find the correct unit for potential difference (voltage,

V

).

A) From Ohm’s Law,

V = IR

.

Units: Volts = Amps

\times

Ohms.

Option ’amp per ohm’ (A/

\Omega

) is incorrect.

B) From Work/Energy,

W = QV

, so

V = \frac{W}{Q}

.

Units: Volts = Joules per Coulomb (J/C).

Option ’coulomb per joule’ (C/J) is incorrect.

C) From Power,

P = \frac{E}{t}

.

Units: Watts = Joules per second (J/s).

This defines the Watt, not the Volt.

D) From Electric Field Strength,

E = \frac{F}{q}

.

Units: N/C. This is the unit for electric field strength, not potential difference. Incorrect.

E) From Electrical Power,

P = IV

, so

V = \frac{P}{I}

.

Units: Volts = Watts per Ampere (W/A). This is correct.

Question 9

1 mark

The right-angled triangle shown has horizontal and vertical sides measuring

(4 + \sqrt{2})

cm and

(2 - \sqrt{2})

cm respectively.

[diagram not to scale]

Calculate the area of the triangle.

A. $(5+3\sqrt{2})\text{cm}^2$
B. $(3-\sqrt{2})\text{cm}^2$
C. $(3+3\sqrt{2})\text{cm}^2$
D. $(5-\sqrt{2})\text{cm}^2$

Answer: B

Worked solution

The area of a right-angled triangle is found using the formula:

\text{Area} = \frac{1}{2} \times \text{base} \times \text{height}

Substitute the given lengths of the base and height:

\text{Area} = \frac{1}{2} (4 + \sqrt{2})(2 - \sqrt{2})

Expand the brackets (using FOIL):

\text{Area} = \frac{1}{2} \left( 4 \times 2 - 4\sqrt{2} + 2\sqrt{2} - \sqrt{2}\sqrt{2} \right)

\text{Area} = \frac{1}{2} (8 - 4\sqrt{2} + 2\sqrt{2} - 2)

Combine the constant terms and the square root terms:

\text{Area} = \frac{1}{2} (6 - 2\sqrt{2})

Distribute the

\frac{1}{2}

to get the final answer:

\text{Area} = 3 - \sqrt{2}

Question 10

1 mark

Two radioactive sources X and Y have half-lives of 4.8 hours and 8.0 hours respectively. Both decay directly to form only stable isotopes.

The activity of a sample of the source X is 320 Bq, and the activity of a sample of the source Y is 480 Bq. The two samples are now combined.

What is the activity of the combination of X and Y 24 hours later?

(An activity of 1 Bq is 1 decay per second.)

A.25 Bq
B.50 Bq
C.55 Bq
D.70 Bq
E.100 Bq
F.140 Bq

Answer: D

Worked solution

Calculate the number of half-lives that occur for each source in 24 hours.

Source X:

- Number of half-lives:

n_X = \frac{24\text{ hours}}{4.8\text{ hours}} = 5

- Final activity:

A_X = \frac{A_{0,X}}{2^{n_X}} = \frac{320}{2^5} = \frac{320}{32} = 10 \text{ Bq}

Source Y:

- Number of half-lives:

n_Y = \frac{24\text{ hours}}{8.0\text{ hours}} = 3

- Final activity:

A_Y = \frac{A_{0,Y}}{2^{n_Y}} = \frac{480}{2^3} = \frac{480}{8} = 60 \text{ Bq}

Total Activity: The total activity is the sum of the individual final activities.

A_{\text{total}} = A_X + A_Y = 10 \text{ Bq} + 60 \text{ Bq} = 70 \text{ Bq}

Question 11

1 mark

A solid sphere of radius r fits inside a hollow cylinder. The cylinder has the same internal diameter and length as the diameter of the sphere.

The volume of a sphere is

\frac{4}{3} \pi r^3

, where r is the radius of the sphere.

What fraction of the space inside the cylinder is taken up by the sphere?

A. $\frac{1}{4}$
B. $\frac{1}{3}$
C. $\frac{1}{2}$
D. $\frac{2}{3}$
E. $\frac{3}{4}$

Answer: D

Worked solution

The volume of the sphere is given as:

V_{\text{sphere}} = \frac{4}{3}\pi r^3

A cylinder that fits the sphere has an internal radius equal to the sphere's radius,

r

, and a height,

h

, equal to the sphere's diameter,

2r

. The volume of the cylinder is calculated as:

V_{\text{cylinder}} = \pi \times (\text{radius})^2 \times (\text{height}) \ = \pi r^2 h \ = \pi r^2 (2r) \ = 2\pi r^3

The fraction of the cylinder's volume taken up by the sphere is the ratio of their volumes:

\text{Fraction} = \frac{V_{\text{sphere}}}{V_{\text{cylinder}}} = \frac{\frac{4}{3}\pi r^3}{2\pi r^3} = \frac{\frac{4}{3}}{2} = \frac{2}{3}

Question 12

1 mark

A cyclist and a bike have a combined mass of 100 kg. The cyclist free-wheels (rolls without pedalling) at a constant speed of

0.80 \text{ms}^{-1}

down a slope where the cyclist descends 1.0 m for each 10 m travelled along the road, as shown in the diagram.

[diagram not to scale]

Calculate the loss in gravitational potential energy as the cyclist loses 100 m in vertical height and hence calculate the total resistive force on the cyclist and bike.

(gravitational field strength =

10 \text{Nkg}^{-1}

)

A.loss in gravitational potential energy/J: 3200, resistive force/N: $32 / \sqrt{101}$
B.loss in gravitational potential energy/J: 3200, resistive force/N: 3.2
C.loss in gravitational potential energy/J: 3200, resistive force/N: $32 / \sqrt{99}$
D.loss in gravitational potential energy/J: 100 000, resistive force/N: $1000 / \sqrt{101}$
E.loss in gravitational potential energy/J: 100 000, resistive force/N: 100
F.loss in gravitational potential energy/J: 100 000, resistive force/N: $1000 / \sqrt{99}$

Answer: E

Worked solution

First, we calculate the loss in gravitational potential energy (GPE) for a vertical height drop of

h = 100 \, \text{m}

.

\text{GPE}_{\text{loss}} = mgh \ = (100 \, \text{kg}) \times (10 \, \text{Nkg}^{-1}) \times (100 \, \text{m}) \ = 100,000 \, \text{J}

The cyclist travels at a constant speed, so there is no change in kinetic energy. This means the GPE lost equals the work done against resistive forces.

\text{Work Done by resistive force} = \text{GPE}_{\text{loss}} = 100,000 \, \text{J}

From the problem, for every

1 \, \text{m}

of vertical descent, the cyclist travels

10 \, \text{m}

along the road. For a

100 \, \text{m}

descent, the distance travelled,

d

, is:

d = 100 \times 10 \, \text{m} = 1000 \, \text{m}

The resistive force,

F

, can now be found:

\text{Work Done} = F \times d \ 100,000 \, \text{J} = F \times 1000 \, \text{m} \ F = \frac{100,000}{1000} \, \text{N} = 100 \, \text{N}

Question 13

1 mark

Which of the expressions below has the largest value for

0 < x < 1

?

A. $\frac{1}{x}$
B. $x^2$
C. $\frac{1}{(1+x)}$
D. $\frac{1}{\sqrt{x}}$
E. $\sqrt{x}$

Answer: A

Worked solution

To find which expression is largest for

0 < x < 1

, we can substitute a test value from this interval. Let’s choose

x = \frac{1}{2}

.

We evaluate each expression:

A:

\frac{1}{x} = \frac{1}{1/2} = 2

B:

x^2 = \left(\frac{1}{2}\right)^2 = \frac{1}{4}

C:

\frac{1}{1 + x} = \frac{1}{1 + \frac{1}{2}} = \frac{1}{\frac{3}{2}} = \frac{2}{3}

D:

\frac{1}{\sqrt{x}} = \frac{1}{\sqrt{1/2}} = \sqrt{2}

E:

\sqrt{x} = \sqrt{\frac{1}{2}} = \frac{1}{\sqrt{2}}

Comparing the results: 2 is the largest value. This corresponds to expression A.

Question 14

1 mark

A ball is thrown vertically upwards and leaves the thrower's hand with a speed of

12 \text{ms}^{-1}

. You may assume that all of the initial kinetic energy of the ball has been converted into gravitational potential energy when the ball reaches its highest point.

What is the height above the thrower's hand to which it rises?

(gravitational field strength =

10 \text{Nkg}^{-1}

)

A.7.2 m
B.14.4 m
C.24 m
D.60 m
E.120 m

Answer: A

Worked solution

By the principle of conservation of energy, the initial kinetic energy (KE) of the ball is converted into gravitational potential energy (GPE) at its highest point.

\text{KE}_{\text{initial}} = \text{GPE}_{\text{final}} \ \frac{1}{2}mv^2 = mgh

The mass

m

cancels out, so we can solve for the height

h

:

h = \frac{v^2}{2g}

We are given the initial speed

v = 12 \, \text{ms}^{-1}

and the gravitational field strength

g = 10 \, \text{Nkg}^{-1}

.

h = \frac{12^2}{2 \times 10} \ h = \frac{144}{20} \ h = 7.2 \, \text{m}

Question 15

1 mark

A shape is formed by drawing a triangle ABC inside the triangle ADE.

BC is parallel to DE.

AB = 4 cm, BC = x cm, DE = x + 3 cm, DB = x - 4 cm

[diagram not to scale]

Calculate the length of DE.

A.5 cm
B.7 cm
C.9 cm
D. $4+2\sqrt{7}$ cm
E. $7+2\sqrt{7}$ cm

Answer: C

Worked solution

Triangles

ABC

and

ADE

are similar because

BC

is parallel to

DE

. The ratio of their corresponding sides is equal.

\frac{DE}{BC} = \frac{AD}{AB}

The length of side

AD

is the sum of

AB

and

DB

:

AD = AB + DB = 4 + (x-4) = x

Substitute the given values into the ratio equation:

\frac{x+3}{x} = \frac{x}{4}

Now, we solve the resulting quadratic equation for

x

:

4(x+3) = x^2 \ 4x + 12 = x^2 \ x^2 - 4x - 12 = 0 \ (x-6)(x+2) = 0

Since length must be a positive value,

x=6

. We calculate the length of

DE

:

DE = x + 3 = 6 + 3 = 9 \, \text{cm}

Question 16

1 mark

A lorry of mass m, and travelling initially at speed v along a horizontal road, is brought to rest by an average horizontal braking force F in time t.

Ignoring any other resistive forces, what distance is travelled by the lorry during this time?

(gravitational field strength =

10 \text{Nkg}^{-1}

)

A. $\frac{F}{mg}$
B. $\frac{mgv}{F}$
C. $\frac{mv^2}{2F}$
D. $\frac{v^2}{2g}$
E.vt
F.2vt

Answer: C

Worked solution

According to the work-energy principle, the work done by the braking force is equal to the change in the lorry's kinetic energy. Since the lorry comes to rest, its final kinetic energy is zero. The work done by the force,

W

, is equal to the initial kinetic energy,

KE_i

.

W = KE_i

\text{Force} \times \text{Distance} = \frac{1}{2}mv^2

Substituting the variables

F

for Force and

d

for Distance:

F \times d = \frac{1}{2}mv^2

To find the distance travelled, we rearrange the equation to solve for

d

:

d = \frac{mv^2}{2F}

Question 17

1 mark

Two variables are connected by the relation:

P \propto \frac{1}{Q^2}

.

Q is increased by 40%.

To the nearest percent, describe the change in P in percentage terms.

A.29% decrease
B.44% decrease
C.49% decrease
D.51% decrease
E.80% decrease
F.96% decrease

Answer: C

Worked solution

The relationship is given by

P \propto \frac{1}{Q^2}

, so we can write

P = \frac{k}{Q^2}

. Let the initial values be

P_1

and

Q_1

. An increase in

Q

by 40% means the new value,

Q_2

, is

1.4 \times Q_1

. The new value for

P

, denoted

P_2

, is:

P_2 = \frac{k}{Q_2^2} = \frac{k}{(1.4 Q_1)^2} = \frac{k}{1.96 Q_1^2}

Since

P_1 = \frac{k}{Q_1^2}

, we can substitute to find

P_2

in terms of

P_1

:

P_2 = \frac{1}{1.96} P_1

The fractional change in

P

is:

\frac{P_2 - P_1}{P_1} = \frac{\frac{1}{1.96}P_1 - P_1}{P_1} = \frac{1}{1.96} - 1 = \frac{1 - 1.96}{1.96} = -\frac{0.96}{1.96}

To find the percentage decrease:

\frac{0.96}{1.96} \times 100\% = \frac{96}{196} \times 100\% = \frac{24}{49} \times 100\%

Approximating the value:

\frac{24}{49} \approx \frac{24}{50} = 0.48 \implies 48\%

Since the denominator is slightly smaller, the result is slightly larger than 48\%.

\frac{24}{49} \approx 0.4897... \approx 49\%

Question 18

1 mark

Nuclide

_R^N X

is an unstable isotope which decays in two stages into nuclide Z as shown:

$_R^N X

\rightarrow

_{R-1}^P Y

\rightarrow

_Q^F Z

What are the values of P and Q?

(Consider only alpha and beta decays.)

A.P: N-4, Q: R+1
B.P: N-4, Q: R-1
C.P: N-4, Q: R-2
D.P: N, Q: R-1
E.P: N, Q: R-2
F.P: N, Q: R-4

Answer: B

Worked solution

The decay process is:

_R^N X \rightarrow \;_{R-2}^{P} Y \rightarrow \;_Q^P Z

\subsection*{Step 1:

X \rightarrow Y

} The atomic number (subscript) decreases by 2 (from

R

to

R-2

). This indicates an alpha decay, which involves the emission of a

^4_2\text{He}

particle. In an alpha decay, the mass number (superscript) decreases by 4.

P = N - 4

\subsection*{Step 2:

Y \rightarrow Z

} The mass number

P

remains constant. This indicates a beta decay, where a neutron becomes a proton, and an electron (

^0_{-1}e

) is emitted. In a beta decay, the atomic number increases by 1. The atomic number of nuclide Y is

R-2

.

Q = (R-2) + 1 = R-1

Question 19

1 mark

Three variables x, y and z are known to be related to each other in the following ways:

x is directly proportional to the square of z.
y is inversely proportional to the cube of z.

Which of the following correctly describes the relationship between x and y?

A.The square of x is directly proportional to the cube of y.
B.The square of x is inversely proportional to the cube of y.
C.The cube of x is directly proportional to the square of y.
D.The cube of x is inversely proportional to the square of y.
E.x is directly proportional to $y^6$ .

Answer: D

Worked solution

The given proportionalities are:

x \propto z^2 \ y \propto \frac{1}{z^3}

We can express a term proportional to

z

from each relation. From the first relation:

\sqrt{x} \propto z

From the second relation:

z^3 \propto \frac{1}{y} \implies z \propto \frac{1}{\sqrt[3]{y}}

Now, we can equate the two expressions that are proportional to

z

:

\sqrt{x} \propto \frac{1}{\sqrt[3]{y}}

To express this relationship with integer powers, we raise both sides to the power of 6, which is the lowest common multiple of the root indices (2 and 3).

(\sqrt{x})^6 \propto \left( \frac{1}{\sqrt[3]{y}} \right)^6 \ x^{(1/2) \times 6} \propto \frac{1}{y^{(1/3) \times 6}} \ x^3 \propto \frac{1}{y^2}

This means the cube of

x

is inversely proportional to the square of

y

.

Question 20

1 mark

A pulse of frequency 100 kHz is emitted from an ultrasound scanner, and is reflected from a foetus 10 cm below the transmitter placed on the mother's abdomen. The speed of sound within the mother's body is

500 \text{ms}^{-1}

.

How long after its emission from the scanner does it take for the pulse to reach the receiver which is adjacent to the transmitter?

A.0.20 ms
B.0.40 ms
C.0.50 ms
D.0.80 ms
E.1.0 ms

Answer: B

Worked solution

The ultrasound pulse travels to the foetus and is reflected back. The total distance is therefore twice the distance to the foetus.

\text{Total distance } (d) = 2 \times 10 \, \text{cm} \ = 20 \, \text{cm} = 0.2 \, \text{m}

The speed of the pulse is given.

\text{Speed } (v) = 500 \, \text{ms}^{-1}

The time taken for the round trip is calculated as:

\text{Time } (t) = \frac{\text{distance}}{\text{speed}} \ = \frac{0.2 \, \text{m}}{500 \, \text{ms}^{-1}} \ = \frac{2}{5000} \, \text{s} \ = 0.0004 \, \text{s}

Converting to milliseconds (ms):

0.0004 \, \text{s} \times 1000 = 0.4 \, \text{ms}

Question 21

1 mark

In the triangle PQR shown below:

[diagram not to scale]

X lies on PR

\angle QXR

is

90^\circ

\frac{QX}{PX} = \frac{1}{6}

\frac{QX}{XR} = \frac{2}{3}

M is the midpoint of PR.

What is

\frac{QX}{MX}

?

A. $\frac{1}{9}$
B. $\frac{5}{12}$
C. $\frac{4}{9}$
D. $\frac{1}{2}$
E. $\frac{5}{6}$

Answer: C

Worked solution

To find the ratio, we can set a reference length. Let

QX = 1

. From the given ratios, we can determine the lengths of

PX

and

XR

:

\frac{QX}{PX} = \frac{1}{6} \implies PX = 6 \times 1 = 6 \ \frac{QX}{XR} = \frac{2}{3} \implies XR = \frac{3}{2} \times 1 = \frac{3}{2}

The total length of the base

PR

is:

PR = PX + XR = 6 + \frac{3}{2} = \frac{15}{2}

M

is the midpoint of

PR

. The distance from P to M is:

PM = \frac{PR}{2} = \frac{15/2}{2} = \frac{15}{4}

The distance

MX

is the difference between

PX

and

PM

:

MX = PX - PM = 6 - \frac{15}{4} = \frac{24 - 15}{4} = \frac{9}{4}

The required ratio is:

\frac{QX}{MX} = \frac{1}{9/4} = \frac{4}{9}

Question 22

1 mark

The diagrams below show either velocity-time or distance-time graphs for four different objects, P, Q, R and S.

Which graph(s) show an object accelerating at

2.4 \text{ms}^{-2}

?

A.P only
B.Q only
C.R only
D.S only
E.P and Q only
F.Q and R only
G.P and S only

Answer: B

Worked solution

Acceleration is the gradient of a velocity-time graph. A straight line on a distance-time graph indicates constant velocity, hence zero acceleration.

Graphs R and S (Distance-Time):

- These graphs show constant velocity (straight line), meaning acceleration is 0.

Graph P (Velocity-Time):

- The gradient represents acceleration.

a = \frac{\Delta v}{\Delta t} = \frac{10 \text{ m/s}}{24 \text{ s}} ≠ 2.4 \text{ ms}^{-2}

Graph Q (Velocity-Time):

- The gradient represents acceleration.

a = \frac{\Delta v}{\Delta t} = \frac{48 \text{ m/s}}{20 \text{ s}} = 2.4 \text{ ms}^{-2}

Only graph Q shows an object accelerating at 2.4 ms

^{-2}

.

Question 23

1 mark

Solve the inequality

x^2 \ge 8 - 2x

A. $x \ge 4$
B. $x \le 2$ and $x \ge -4$
C. $x \ge -2$ and $x \le 4$
D. $x \ge 2$ or $x \le -4$

Answer: D

Worked solution

To solve the inequality, first rearrange it into the standard quadratic form with zero on one side.

x^2 \geq 8 - 2x \ x^2 + 2x - 8 \geq 0

Next, factorise the quadratic expression.

(x+4)(x-2) \geq 0

For the product to be non-negative, either both terms are non-negative, or both are non-positive. Case 1: Both terms are non-negative.

(x+4) \geq 0 \text{ and } (x-2) \geq 0 \implies x \geq 2

Case 2: Both terms are non-positive.

(x+4) \leq 0 \text{ and } (x-2) \leq 0 \implies x \leq -4

Combining these two cases, the solution is

x \geq 2

or

x \leq -4

.

Question 24

1 mark

Which one of the following statements about nuclear physics is true?

A.The process of emission of a gamma ray from a nucleus is called nuclear fission.
B.The half-life of a radioactive substance is half the time taken for its nuclei to decay.
C.The number of neutrons in a nucleus is its atomic number (proton number) minus its mass number.
D.When a nucleus emits a beta particle, there is no change in the number of particles it contains.
E.When a nucleus emits an alpha particle, one of its neutrons becomes a proton plus an electron.

Answer: D

Worked solution

We can determine the correct statement by a process of elimination, analysing the definitions of key concepts in nuclear physics.

A: This is incorrect. The emission of a gamma ray is a de-excitation process where a nucleus loses energy. Nuclear fission is the splitting of a heavy nucleus into two or more lighter nuclei.

B: This is incorrect. The half-life is the time it takes for half of the radioactive nuclei in a sample to undergo decay.

C: This is incorrect. The number of neutrons is the mass number (

A

) minus the atomic number (

Z

). The statement reverses this relationship.

E: This is incorrect. This statement describes beta decay (

n \to p + e^- + \bar{ u}_e

). Alpha decay is the emission of an alpha particle (a helium nucleus).

D: This is correct. In beta decay, a neutron converts into a proton or vice-versa. This changes the neutron and proton numbers individually, but the total number of nucleons (the mass number) remains constant.

Question 25

1 mark

The total surface area of a cylinder, measured in square centimetres, is numerically the same as its volume, measured in cubic centimetres.

The radius of the cylinder is r cm, the height is h cm.

Express h in terms of r.

A. $h = \frac{2r}{r-2}$
B. $h = \frac{2r}{r+2}$
C. $h = r+2$
D. $h = r-2$
E. $h = 2r(r-2)$

Answer: A

Worked solution

The total surface area of a cylinder is the sum of the area of the two circular ends and the curved side.

A = 2\pi r^2 + 2\pi rh

The volume of a cylinder is given by:

V = \pi r^2 h

The question states that the surface area and volume are numerically equal.

2\pi r^2 + 2\pi rh = \pi r^2 h

Divide the equation by the common factor

\pi

.

2r^2 + 2rh = r^2h

Rearrange the equation to isolate terms involving

h

.

2r^2 = r^2h - 2rh

Factor out

h

from the right-hand side.

2r^2 = h(r^2 - 2r)

Solve for

h

and simplify the expression.

h = \frac{2r^2}{r^2 - 2r} = \frac{2r^2}{r(r-2)} = \frac{2r}{r-2}

Question 26

1 mark

Two resistors with resistance

R_1

ohms and

R_2

ohms are connected in series with a battery that has a voltage V across its terminals.

Which formula gives the power dissipated by the resistor with resistance

R_1

ohms?

A. $\frac{VR_1}{R_1 + R_2}$
B. $\frac{V^2R_1}{R_1 + R_2}$
C. $\frac{VR_1^2}{(R_1 + R_2)^2}$
D. $\frac{V^2R_1}{(R_1 + R_2)^2}$
E. $\frac{V^2R_1^2}{(R_1 + R_2)^2}$
F. $\frac{V^2R_2^2}{(R_1 + R_2)^2}$

Answer: D

Worked solution

For resistors in series, the total resistance is the sum of individual resistances.

R_{\text{total}} = R_{1} + R_{2}

The total current

I

flowing through the circuit is given by Ohm's law:

I = \frac{V}{R_{\text{total}}} = \frac{V}{R_{1} + R_{2}}

This current flows through both resistors. The voltage

V_{1}

across resistor

R_{1}

is:

V_{1} = I \times R_{1} = \frac{V R_{1}}{R_{1} + R_{2}}

The power

P_{1}

dissipated by resistor

R_{1}

is given by the formula

P = \frac{V^2}{R}

.

P_{1} = \frac{V_{1}^2}{R_{1}}

Substituting the expression for

V_{1}

:

P_{1} = \frac{\left(\frac{V R_{1}}{R_{1} + R_{2}}\right)^2}{R_{1}} = \frac{V^2 R_{1}^2}{(R_{1} + R_{2})^2} \times \frac{1}{R_{1}}

P_{1} = \frac{V^{2} R_{1}}{(R_{1} + R_{2})^{2}}

Question 27

1 mark

The square PQRS is positioned so that its vertices are at the points with coordinates: (1, 1), (-1, 1), (-1, -1) and (1, -1).

The square is rotated clockwise through

90^\circ

about the origin and then reflected in the line

y = x

.

Which transformation will return the square to its original orientation?

A.A reflection in the x-axis.
B.A reflection in the y-axis.
C.A reflection in the line $y = -x$ .
D.A rotation of $90^\circ$ clockwise about the origin.
E.A rotation of $90^\circ$ anticlockwise about the origin.

Answer: B

Worked solution

Let’s track the transformation of a general point

(x, y)

on the square.

Step 1: Rotation

A clockwise rotation of 90° about the origin maps a point

(x, y)

to

(y, -x)

.

Step 2: Reflection

A reflection in the line

y = x

maps a point

(x', y')

to

(y', x')

. Applying this to our transformed point from Step 1,

(y, -x)

, we get

(-x, y)

.

Combined Transformation

The overall transformation is

(x, y) \to (-x, y)

. This is the definition of a reflection in the

y

-axis.

To return the square to its original state, we must apply the inverse of this transformation. A reflection is its own inverse, so the required transformation is another reflection in the

y

-axis.

Question 28

1 mark

A sound wave is produced by a loudspeaker cone, which creates pulses of pressure by moving back and forth between two points X and Y as shown in the diagram.

[diagram not to scale]

The distance between points X and Y is 5.0 mm and the loudspeaker produces pulses of high pressure every 0.20 milliseconds.

The following statements about the sound wave produced are made:

P It has a speed of

25 \text{ms}^{-1}

.
Q It has an amplitude of 5.0 mm.
R It has a wavelength of 5.5 mm.
S It has a frequency of 5.0 kHz.

Which of these statements can be correctly deduced from the information given?

A.P only
B.S only
C.P and Q only
D.P and R only
E.Q and S only
F.R and S only
G.P, R and S only

Answer: B

Worked solution

We evaluate each statement based on the information given.

\\boldsymbol{P\ (Speed)}

: The speed of the sound wave cannot be determined from the information provided, as it depends on the medium.

\\boldsymbol{Q\ (Amplitude)}

: The cone moves a total distance of 5.0 mm. The amplitude is the maximum displacement from the central equilibrium position, which is half of the total distance.

Amplitude = \frac{5.0\ \text{mm}}{2} = 2.5\ \text{mm}

Thus, statement Q is incorrect.

\\boldsymbol{R\ (Wavelength)}

: The wavelength

\lambda

is given by the formula

c = f \lambda

. Since the wave speed

c

is unknown, the wavelength cannot be calculated.

\\boldsymbol{S\ (Frequency)}

: The time period

T

is given as 0.20 milliseconds.

T = 0.20\ \text{ms} = 0.20 \times 10^{-3}\ \text{s}

Frequency

f

is the reciprocal of the time period.

f = \frac{1}{T} = \frac{1}{0.20 \times 10^{-3}\ \text{s}} = 5000\ \text{Hz} = 5.0\ \text{kHz}

Statement S can be correctly deduced.

Only statement S can be deduced.

Question 29

1 mark

Which one of the following is a simplification of

\frac{x^2 - 4}{x^2 - 2x}

where

x ≠ 2

and

x ≠ 0

?

A. $\frac{x-4}{x-2}$
B. $\frac{x-2}{x}$
C. $\frac{2}{x}$
D. $\frac{x+2}{x}$
E. $\frac{x+2}{x+1}$

Answer: D

Worked solution

To simplify the algebraic fraction, we factorise the numerator and the denominator.

\frac{x^2 - 4}{x^2 - 2x} = \frac{(x+2)(x-2)}{x(x-2)}

The numerator is factorised as a difference of two squares, and the denominator by taking out the common factor of

x

. Since the problem states

x ≠ 2

, the term

(x-2)

is non-zero and can be cancelled from the top and bottom.

= \frac{x+2}{x}

Question 30

1 mark

A car and driver, of total mass 1000 kg, travel with uniform acceleration from rest along a horizontal road. The car's engine produces a force of 3.0 kN. After a time of 5.0 s the car reaches a speed of

10 \text{ms}^{-1}

. Consider the resistive forces acting on the car to be constant.

What is the acceleration of the car

a \text{ms}^{-2}

and what is the total of the resistive forces

f

kN acting on the car?

A.a = 2.0; f = 1.0
B.a = 2.0; f = 2.0
C.a = 2.0; f = 3.0
D.a = 3.0; f = 1.0
E.a = 3.0; f = 2.0

Answer: A

Worked solution

First, we find the acceleration,

a

, of the car.

a = \frac{\Delta v}{t} = \frac{10 \text{ m s}^{-1}}{5.0 \text{ s}} = 2.0 \text{ m s}^{-2}

Next, we use Newton's second law,

F_{net} = ma

, to find the resistive force,

f

. The net force is the difference between the engine's driving force and the resistive force. The driving force is

3.0 \text{ kN} = 3000 \text{ N}

. The car's mass is

1000 \text{ kg}

.

F_{driving} - f = ma

3000 - f = 1000 \times 2.0

3000 - f = 2000

Solving for

f

:

f = 3000 - 2000 = 1000 \text{ N}

Converting the resistive force to kilonewtons:

f = 1.0 \text{ kN}

The acceleration is

a = 2.0 \text{ m s}^{-2}

and the resistive force is

f = 1.0 \text{ kN}

.

Question 31

1 mark

Given that

a^x b^{2x} c^{3x} = 2

, where a, b, and c are positive real numbers, then

x =

A. $\log_{10}(\frac{2}{a+2b+3c})$
B. $\frac{\log_{10} 2}{\log_{10} (a+2b+3c)}$
C. $\frac{2}{\log_{10} (a+2b+3c)}$
D. $\frac{2}{a+2b+3c}$
E. $\log_{10}(\frac{2}{ab^2c^3})$
F. $\frac{\log_{10} 2}{\log_{10} (ab^2c^3)}$
G. $\frac{2}{\log_{10} (ab^2c^3)}$
H. $\frac{2}{ab^2c^3}$

Answer: F

Worked solution

Given the equation

a^x b^{2x} c^{3x} = 2

.

Take the logarithm base 10 of both sides:

\log_{10}(a^x b^{2x} c^{3x}) = \log_{10} 2

Apply the logarithm properties

\log(MNP) = \log M + \log N + \log P

and

\log(M^k) = k \log M

:

x \log_{10} a + 2x \log_{10} b + 3x \log_{10} c = \log_{10} 2

Factor out

x

:

x(\log_{10} a + 2 \log_{10} b + 3 \log_{10} c) = \log_{10} 2

Combine the terms in the parenthesis using logarithm rules:

x \log_{10} (ab^2 c^3) = \log_{10} 2

Solve for

x

:

x = \frac{\log_{10} 2}{\log_{10} (ab^2 c^3)}

Question 32

1 mark

Particle P has a fixed mass of 2 kg and particle Q has a fixed mass of 5 kg.

The two particles are moving in opposite directions along a straight line on a smooth plane.

Particle P has a speed of

3 \text{ms}^{-1}

and particle Q has a speed of

r \text{ms}^{-1}

.

The particles collide directly. After the collision the direction of each particle is reversed.

The speed of P is now

1 \text{ms}^{-1}

and the speed of Q is halved.

What is the value of r?

A. $\frac{8}{15}$
B. $\frac{14}{15}$
C. $\frac{16}{15}$
D. $\frac{8}{3}$
E. $\frac{16}{5}$

Answer: C

Worked solution

Let the initial direction of particle P be positive. According to the principle of conservation of linear momentum, the total momentum before the collision equals the total momentum after.

Momentum Before Collision

Total Momentum

= m_P u_P + m_Q u_Q

= (2)(+3) + (5)(-r)

= 6 - 5r

Momentum After Collision

(Directions are reversed)

Total Momentum

= m_P v_P + m_Q v_Q

= (2)(-1) + (5) \left(+ \frac{r}{2} \right)

= -2 + \frac{5r}{2}

Equating Momenta

6 - 5r = -2 + \frac{5r}{2}

8 = 5r + \frac{5r}{2}

8 = \frac{10r + 5r}{2}

8 = \frac{15r}{2}

16 = 15r

r = \frac{16}{15}

Question 33

1 mark

Which one of the following numbers is largest in value?

(All angles are given in radians.)

A. $\tan(\frac{3\pi}{4})$
B. $\log_{10} 100$
C. $\sin^{10}(\frac{\pi}{2})$
D. $\log_2 10$
E. $(\sqrt{2}-1)^{10}$

Answer: D

Worked solution

We evaluate each option to determine the largest value: \begin{itemize} \item[A:]

\tan\left(\frac{3\pi}{4}\right) = -1

\item[B:]

\log_{10}(100) = \log_{10}(10^2) = 2

\item[C:]

\sin^{10}\left(\frac{\pi}{2}\right) = (1)^{10} = 1

\item[D:] For

\log_{2}(10)

, we know

2^3 = 8

and

2^4 = 16

. Since

8 < 10 < 16

, the value must be between 3 and 4. \item[E:] For

(\sqrt{2}-1)^{10}

, the base

\sqrt{2}-1

is a positive number less than 1. Raising it to the power of 10 results in a very small positive number. \end{itemize} Comparing the results, the value from option D is the largest.

Question 34

1 mark

A parachutist falls from an aircraft and reaches a terminal velocity. After a while he opens his parachute and reaches a new (lower) terminal velocity.

Which graph shows how the total air resistance (drag) force acting on him and the parachute varies with time during the fall?

A.Graph A
B.Graph B
C.Graph C
D.Graph D
E.Graph E

Answer: A

Worked solution

The air resistance (drag) is a resistive force, so its magnitude cannot be negative. This fact eliminates options B and D. When an object travels at a constant terminal velocity, the net force on it is zero. This implies that the drag force is equal in magnitude to the object's weight (

F_{drag} = W

). This is true for the first terminal velocity (before the parachute opens) and the second, lower terminal velocity (after it opens). Therefore, the drag force during both periods of terminal velocity must be the same, meaning the horizontal plateaus on the graph must be at the same height. Graph A is the only option that satisfies these conditions. It shows the drag force increasing to a plateau, spiking when the parachute opens, and then decreasing back to the same plateau level.

Question 35

1 mark

The sum of the roots of the equation

2^{2x} - 8 \times 2^x + 15 = 0

is

A.3
B.8
C. $2\log_{10} 2$
D. $\log_{10}(\frac{15}{4})$
E. $\frac{\log_{10} 15}{\log_{10} 2}$

Answer: E

Worked solution

The equation is

2^{2x} - 8 \cdot 2^x + 15 = 0

. Let

y=2^x

. The equation becomes a quadratic in

y

:

y^2 - 8y + 15 = 0

Factorising the quadratic expression:

(y-3)(y-5) = 0

This gives two solutions for

y

:

y=3 \quad \text{or} \quad y=5

Substitute back

y=2^x

to find the roots for

x

:

2^x = 3 \quad \text{or} \quad 2^x = 5

Solving for the two roots (

x_1, x_2

) by taking logarithms:

x_1 = \log_2 3 \quad \text{and} \quad x_2 = \log_2 5

The sum of the roots is

x_1 + x_2

:

\text{Sum} = \log_2 3 + \log_2 5

Using the logarithm property

\log_b m + \log_b n = \log_b(mn)

:

\text{Sum} = \log_2 (3 \times 5) = \log_2 15

Using the change of base formula,

\log_b a = \frac{\log_{c} a}{\log_{c} b}

, to convert to base 10:

\text{Sum} = \frac{\log_{10} 15}{\log_{10} 2}

Question 36

1 mark

A car is accelerated from rest along a horizontal road by a constant thrust force produced by the engine. The car eventually reaches a terminal velocity.

The graphs below show the variation with time of three quantities (X, Y and Z) for the car:

Which line in the table could correctly identify the quantities X, Y and Z?

A.X: acceleration, Y: air resistance (drag force), Z: kinetic energy
B.X: acceleration, Y: mass of car, Z: weight of car
C.X: (gravitational) potential energy, Y: velocity, Z: kinetic energy
D.X: (gravitational) potential energy, Y: air resistance (drag force), Z: weight of car
E.X: resultant force, Y: mass of car, Z: kinetic energy
F.X: resultant force, Y: velocity, Z: weight of car

Answer: F

Worked solution

The car moves on a horizontal road, so its mass, weight, and gravitational potential energy are all constant.

- Graph Z shows a constant quantity. This can be the weight of the car. This eliminates options A, C, and E, which suggest Z is kinetic energy. Kinetic energy increases from zero as the car accelerates, so it is not constant.

- Graph Y shows a non-constant quantity. This eliminates option B, where Y is the mass of the car.

- Graph X shows a non-constant quantity. This eliminates option D, where X is gravitational potential energy.

By elimination, option F is the only choice that is consistent with the physics of the situation.

Question 37

1 mark

For any real numbers a, b, and c where

a \ge b

, consider these three statements:

1.

-b \ge -a

2.

a^2 + b^2 \ge 2ab

3.

ac \ge bc

Which of the above statements must be true?

A.none of them
B.1 only
C.2 only
D.3 only
E.1 and 2 only
F.1 and 3 only
G.2 and 3 only
H.1, 2 and 3

Answer: E

Worked solution

We are given that

a, b, c

are real numbers and

a \geq b

. We must evaluate the three statements. \textbf{Statement 1:

-b \geq -a

} We start with the given inequality:

a \geq b

When we multiply an inequality by a negative number, such as

-1

, the inequality sign is reversed.

(-1) \times a \leq (-1) \times b

-a \leq -b

This is equivalent to

-b \geq -a

. Thus, statement 1 is

\textbf{true}

. \textbf{Statement 2:

a^2 + b^2 \geq 2ab

} The square of any real number is non-negative. Therefore:

(a - b)^2 \geq 0

Expanding the left side gives:

a^2 - 2ab + b^2 \geq 0

Rearranging the terms:

a^2 + b^2 \geq 2ab

This is true for all real numbers

a

and

b

, regardless of whether

a \geq b

. Thus, statement 2 is

\textbf{true}

. \textbf{Statement 3:

ac \geq bc

} Starting with

a \geq b

, the outcome of multiplying by

c

depends on its sign. \begin{itemize} \item If

c > 0

, then

ac \geq bc

(inequality holds). \item If

c < 0

, then

ac \leq bc

(inequality reverses). \end{itemize} Since

c

can be any real number, this statement is not always true. Therefore, only statements 1 and 2 must be true.

Question 38

1 mark

A heavy block of stone rests on a rough, horizontal surface.

The block is subject to a horizontal force that increases from zero at a constant rate.

Assume that the coefficient of friction is greater than zero and that its value is independent of whether or not the block is moving.

What happens to the block of stone?

(Assume air resistance is negligible.)

A.It moves forwards immediately and accelerates forwards with a constant acceleration.
B.It remains stationary at first and then accelerates forwards with a constant acceleration.
C.It remains stationary at first and then accelerates forwards with an increasing acceleration.
D.It moves forwards immediately with a constant velocity.
E.It remains stationary at first and then moves forwards with a constant velocity.

Answer: C

Worked solution

Initially, the block is stationary. The static frictional force balances the applied force. The block will only start to move when the applied force exceeds the maximum static friction, which is a fixed value. Let the applied force be

F_{app}

and the constant kinetic frictional force be

F_f

. The applied force increases linearly with time. Once the block is moving, the net force is:

F_{net} = F_{app} - F_f

By Newton's second law,

F_{net} = ma

:

a = \frac{F_{net}}{m} = \frac{F_{app} - F_f}{m}

Since the applied force

F_{app}

is continuously increasing and

F_f

is constant, the net force

F_{net}

also increases. Thus, the acceleration

a

must also increase. The block remains stationary at first, then moves with an increasing acceleration.

Question 39

1 mark

The sequence

a_n

is given by the rule:

a_1 = 2

a_{n+1} = a_n + (-1)^n

for

n \ge 1

What is

\sum_{n=1}^{100} a_n

?

A.150
B.250
C.-4750
D.5150
E. $4(1-(\frac{1}{2})^{100})$
F. $4((\frac{3}{2})^{100}-1)$

Answer: A

Worked solution

The sequence is defined by the recurrence relation:

a_1 = 2 \ a_{n+1} = a_n + (-1)^n \quad \text{for } n \ge 1

We calculate the first few terms to find a pattern:

a_1 = 2 \ a_2 = a_1 + (-1)^1 = 2 - 1 = 1 \ a_3 = a_2 + (-1)^2 = 1 + 1 = 2 \ a_4 = a_3 + (-1)^3 = 2 - 1 = 1 \ a_5 = a_4 + (-1)^4 = 1 + 1 = 2

The sequence is

\{2, 1, 2, 1, \dots\}

. For odd terms

a_n=2

, and for even terms

a_n=1

. The sum required is

\sum_{n=1}^{100} a_n

. This is the sum of 100 terms. The sum can be grouped into 50 pairs:

\sum_{n=1}^{100} a_n = (a_1+a_2) + (a_3+a_4) + \dots + (a_{99}+a_{100})

Each pair consists of an odd term (value 2) and an even term (value 1).

\text{Sum of each pair} = 2 + 1 = 3

There are 50 such pairs.

\text{Total Sum} = 50 \times 3 = 150

Question 40

1 mark

A white billiard ball of mass 0.20 kg is travelling horizontally at

3.0 \text{ms}^{-1}

and hits a red billiard ball of the same mass which is at rest. After the collision the white ball continues in the same direction with a speed of

1.0 \text{ms}^{-1}

.

What is the speed of the red ball immediately after the collision?

A. $1.0 \text{ms}^{-1}$
B. $1.5 \text{ms}^{-1}$
C. $2.0 \text{ms}^{-1}$
D. $2.5 \text{ms}^{-1}$
E. $3.0 \text{ms}^{-1}$

Answer: C

Worked solution

This problem is solved using the principle of conservation of linear momentum, where the total momentum before the collision equals the total momentum after.
ewline Let the mass of each ball be

m=0.20\,\text{kg}

and the final velocity of the red ball be

v

.
ewline The initial momentum is calculated from the white ball, as the red ball is at rest.

\text{Momentum before} = (0.20\,\text{kg} \times 3.0\,\text{ms}^{-1}) + (0.20\,\text{kg} \times 0\,\text{ms}^{-1}) \ = 0.6\,\text{Ns}

The final momentum is the sum of the momenta of both balls.

\text{Momentum after} = (0.20\,\text{kg} \times 1.0\,\text{ms}^{-1}) + (0.20\,\text{kg} \times v) \ = 0.2 + 0.2v

Equating the momentum before and after the collision:

0.6 = 0.2 + 0.2v \ 0.4 = 0.2v \ v = \frac{0.4}{0.2} \ v = 2.0\,\text{ms}^{-1}

Question 41

1 mark

How many real roots does the equation

x^4 - 4x^3 + 4x^2 - 1 = 0

have?

A.0
B.1
C.2
D.3
E.4

Answer: C

Worked solution

To find the number of real roots for

y = x^4 - 4x^3 + 4x^2 - 10

, we locate its turning points. First, we find the derivative:

\frac{dy}{dx} = 4x^3 - 12x^2 + 8x

Set the derivative to zero to find the stationary points:

4x^3 - 12x^2 + 8x = 0

4x(x^2 - 3x + 2) = 0

4x(x-1)(x-2) = 0

The stationary points occur at

x=0

,

x=1

, and

x=2

. Next, evaluate the function at these x-values: \begin{itemize} \item At

x=0

,

y = -10

. \item At

x=1

,

y = 1^4 - 4(1)^3 + 4(1)^2 - 10 = -9

. \item At

x=2

,

y = 2^4 - 4(2)^3 + 4(2)^2 - 10 = 16 - 32 + 16 - 10 = -10

. \end{itemize} The turning points are

(0,-10)

,

(1,-9)

, and

(2,-10)

. Since the function is a positive quartic and all turning points are below the x-axis, the graph must cross the x-axis exactly twice.

Question 42

1 mark

A certain planet has no atmosphere. The planet has a gravitational field strength at its surface of

g_p \text{Nkg}^{-1}

; this value is considered constant for this question. A rock is projected vertically upwards from the surface of the planet at an initial speed of

20 \text{ms}^{-1}

. The rock reaches a maximum height h metres.

Which option shows a possible correct pair of values for

g_p

and h?

(Consider only gravitational forces.)

A. $g_p = 5.0 ; h = 4.0$
B. $g_p = 5.0 ; h = 40$
C. $g_p = 10 ; h = 2.0$
D. $g_p = 20 ; h = 1.0$
E. $g_p = 20 ; h = 20$

Answer: B

Worked solution

By the principle of conservation of energy, the initial kinetic energy is converted into gravitational potential energy at the maximum height

h

.

\frac{1}{2}mv^2 = mg_ph

The mass

m

cancels out. We can rearrange for the product

g_ph

:

g_ph = \frac{v^2}{2}

Given the initial speed

v = 20 \, \text{ms}^{-1}

:

g_ph = \frac{20^2}{2} = \frac{400}{2} = 200

We check the options to find which pair of values for

g_p

and

h

gives a product of 200. For option B:

g_p = 5.0

and

h = 40

.

g_ph = 5.0 \times 40 = 200

This is the only option that satisfies the condition.

Question 43

1 mark

A box is a hollow pyramid. The base of the box is a square with sides 10 cm and all the slant edges of the box are 12 cm long.

[diagram not to scale]

What is the angle made by the slant edge TP with the base PQRS?

A. $\sin^{-1} \frac{2\sqrt{5}}{12}$
B. $\sin^{-1} \frac{5}{12}$
C. $\sin^{-1} \frac{5\sqrt{2}}{12}$
D. $\cos^{-1} \frac{2\sqrt{5}}{12}$
E. $\cos^{-1} \frac{5}{12}$
F. $\cos^{-1} \frac{5\sqrt{2}}{12}$

Answer: F

Worked solution

Let the center of the square base be M. The required angle,

x

, is the angle between the slant edge TP and the line PM, where PM is the projection of TP onto the base. This forms a right-angled triangle TMP, with the right angle at M. The hypotenuse is the slant edge,

TP = 12 \, \text{cm}

. The side PM is half the length of the diagonal of the square base. First, find the length of the diagonal of the base (e.g., PR) using Pythagoras' Theorem:

PR^2 = 10^2 + 10^2 = 200

PR = \sqrt{200} = 10\sqrt{2} \, \text{cm}

The distance from a corner to the center is half the diagonal:

PM = \frac{1}{2} \times PR = \frac{10\sqrt{2}}{2} = 5\sqrt{2} \, \text{cm}

In the right-angled triangle TMP, we can find the angle

x

using cosine:

\cos(x) = \frac{\text{Adjacent}}{\text{Hypotenuse}} = \frac{PM}{TP}

\cos(x) = \frac{5\sqrt{2}}{12}

x = \cos^{-1}\left(\frac{5\sqrt{2}}{12}\right)

Question 44

1 mark

A ball is dropped from a height of 16 m on to a horizontal surface, and on each bounce loses 50% of its kinetic energy.

After which bounce will the maximum height of the rebound fall to less than 160 cm for the first time?

(Assume air resistance is negligible, and the only external force acting on the ball while not in contact with the surface is gravity.)

A.4th
B.5th
C.6th
D.7th
E.8th

Answer: A

Worked solution

The gravitational potential energy (GPE) is directly proportional to height (

E_p = mgh

). A 50% loss of kinetic energy during a bounce means the rebound GPE, and thus the rebound height, is 50% of the pre-bounce height.

h_{after} = 0.5 \times h_{before}

The initial height is 16 m. The target height is less than 160 cm = 1.6 m. We track the maximum height after each bounce:

- After bounce 1:

h_1 = 0.5 \times 16 \text{ m} = 8 \text{ m}

- After bounce 2:

h_2 = 0.5 \times 8 \text{ m} = 4 \text{ m}

- After bounce 3:

h_3 = 0.5 \times 4 \text{ m} = 2 \text{ m}

- After bounce 4:

h_4 = 0.5 \times 2 \text{ m} = 1 \text{ m}

The height after the 4th bounce (1 m) is the first to be less than 1.6 m.

Question 45

1 mark

The variables x and y and the constants a and b are real and positive. The variables x and y are related.

A graph of

\log y

against

\log x

is drawn.

For which one of the following relationships will this graph be a straight line?

A. $y^b = a^x$
B. $y = ab^x$
C. $y^2 = a + x^b$
D. $y = ax^b$
E. $y^x = a^b$

Answer: D

Worked solution

A graph of

\log y

against

\log x

is a straight line if the equation can be expressed in the linear form

Y = mX + c

, where

Y = \log y

and

X = \log x

. The target form is

\log y = m\log x + c

. Let's take logarithms of each option: \begin{itemize} \item[A:]

y^6 = a^x \implies 6\log y = x\log a

. Linear in

x

, not

\log x

. \item[B:]

y = ab^x \implies \log y = \log a + x\log b

. Linear in

x

, not

\log x

. \item[C:]

y^2 = a + x^b \implies 2\log y = \log(a+x^b)

. Not a linear form. \item[D:]

y = ax^b \implies \log y = \log a + b\log x

. This is in the form

\log y = (b)\log x + (\log a)

, which is linear. \item[E:]

y^x = a^b \implies x\log y = b\log a

. Not a linear form. \end{itemize} Only option D gives a straight line relationship between

\log y

and

\log x

.

Question 46

1 mark

A man of weight 600 N stands on a set of accurate weighing scales in a moving elevator (lift). The reading on the scales is 480 N.

Which statement correctly describes the motion of the elevator?

A.The elevator is moving downwards with constant speed.
B.The elevator is moving downwards with decreasing speed.
C.The elevator is moving upwards with increasing speed.
D.The elevator is moving upwards with constant speed.
E.The elevator is moving upwards with decreasing speed.

Answer: E

Worked solution

Let

W

be the true weight of the man and

R

be the reading on the scales, which is the normal reaction force.

W = 600 \text{ N} \quad (\text{Force acting downwards})

R = 480 \text{ N} \quad (\text{Force acting upwards})

The net force,

F_{net}

, on the man is the resultant of these two forces. Using Newton’s Second Law (

F_{net} = ma

), and taking the upward direction as positive:

F_{net} = R - W

ma = 480 \text{ N} - 600 \text{ N}

ma = -120 \text{ N}

The negative sign indicates that the net force and hence the acceleration,

a

, are in the downward direction.

An object has a downward acceleration if it is either moving downwards with increasing speed, or moving upwards with decreasing speed.

Let’s check the options:

- A, D: Constant speed means

a = 0

. Incorrect.

- B: Decreasing downward speed means acceleration is upwards. Incorrect.

- C: Increasing upward speed means acceleration is upwards. Incorrect.

- E: Decreasing upward speed means acceleration is downwards. Correct.

Question 47

1 mark

For what values of the non-zero real number a does the quadratic equation

ax^2 + (a-2)x = 2

have distinct real roots?

A.all values of a
B.a = -2
C.a > -2
D.a ≠ -2
E.no values of a

Answer: D

Worked solution

For a quadratic equation to have distinct real roots, its discriminant (

b^2 - 4ac

) must be greater than zero. First, rearrange the equation

ax^2 + (a-2)x = 2

into the standard form

ax^2+bx+c=0

:

ax^2 + (a-2)x - 2 = 0

Here, the coefficients are

a=a

,

b=(a-2)

, and

c=-2

. Now, set the discriminant to be greater than zero:

(a-2)^2 - 4(a)(-2) > 0 \ (a^2 - 4a + 4) + 8a > 0 \ a^2 + 4a + 4 > 0 \ (a+2)^2 > 0

The square of a real number,

(a+2)^2

, is always positive, except when it is zero. This occurs when

a+2=0

, which means

a=-2

. Therefore, the inequality is true for all real values of

a

except for

a=-2

.

Question 48

1 mark

The track for a tram is straight and horizontal. A tram is travelling along the track at a velocity of

12.0 \text{ms}^{-1}

when the brakes are applied. Because of this, the tram decelerates to rest at a constant rate of

1.50 \text{ms}^{-2}

.

What is the distance travelled by the tram over the total time for which it is decelerating?

A.18.0 m
B.48.0 m
C.96.0 m
D.108 m
E.216 m

Answer: B

Worked solution

This is a problem of kinematics with constant acceleration. We use the equation of motion:

v^2 = u^2 + 2as

Given values are:

- Initial velocity,

u = 12.0 \ ms^{-1}

- Final velocity,

v = 0 \ ms^{-1}

(since the tram comes to rest)

- Acceleration,

a = -1.50 \ ms^{-2}

(a deceleration)

We need to find the distance,

s

.

Substitute the values into the equation:

0^2 = (12.0)^2 + 2(-1.50)s

0 = 144 - 3.0 s

3s = 144

s = \frac{144}{3}

s = 48 \ m

Question 49

1 mark

A triangle is to be drawn with sides that are integer lengths in centimetres, and a total perimeter of 12 cm.

How many different (non-congruent) triangles can be drawn?

A.1
B.2
C.3
D.10
E.12

Answer: C

Worked solution

Let the integer side lengths of the triangle be

a, b,

and

c

. The perimeter is 12, so

a + b + c = 12

.

The Triangle Inequality Theorem requires that the sum of the lengths of any two sides must be greater than the length of the third side. This implies that no single side can be greater than or equal to half the perimeter. Thus, each side must be less than 6 cm.

We list the possible combinations of integers

\{a, b, c\}

that sum to 12 and satisfy this condition:

-

\{4, 4, 4\}

: An equilateral triangle.

Check:

4 + 4 > 4

. This is valid.

-

\{5, 4, 3\}

: A scalene triangle.

Check:

3 + 4 > 5

. This is valid.

-

\{5, 5, 2\}

: An isosceles triangle.

Check:

2 + 5 > 5

. This is valid.

Any other set of integers summing to 12, such as

\{6, 5, 1\}

, will violate the triangle inequality (

1 + 5 ot> 6

). Thus, there are 3 possible non-congruent triangles.

Question 50

1 mark

A particle of weight 5 N is held in position by two light ropes.

One of the ropes makes an angle of

60^\circ

with the upward vertical, the other is horizontal.

What is the tension in the horizontal rope?

A. $1.25\sqrt{3}$ N
B.5 N
C. $5\sqrt{3}$ N
D.10 N
E. $10\sqrt{3}$ N

Answer: C

Worked solution

The particle is held in position, which means it is in equilibrium. The net force is zero, so the horizontal and vertical components of the forces must balance. Let

H

be the tension in the horizontal rope and

F

be the tension in the angled rope. For vertical equilibrium, the upward component of

F

must balance the downward weight:

F \cos(60^\circ) = 5

F \times \frac{1}{2} = 5 \implies F = 10 \, \text{N}

For horizontal equilibrium, the tension

H

must balance the horizontal component of

F

:

H = F \sin(60^\circ)

Substituting the value of

F

:

H = 10 \times \frac{\sqrt{3}}{2}

H = 5\sqrt{3} \, \text{N}

Question 51

1 mark

The triangle PQR has a right angle at R.

The length of PQ is 4 cm, correct to the nearest centimetre.

The length of PR is 2 cm, correct to the nearest centimetre.

Find the minimum possible length, in centimetres, of QR.

A. $\sqrt{6} - \frac{1}{2}$
B. $2\sqrt{3} - \frac{1}{2}$
C. $2\sqrt{5} - \frac{1}{2}$
D. $2\sqrt{5}$
E. $2\sqrt{3}$
F. $\sqrt{6}$

Answer: F

Worked solution

The triangle PQR is right-angled at R. By Pythagoras' theorem:

PQ^2 = PR^2 + QR^2 \implies QR = \sqrt{PQ^2 - PR^2}

The lengths are given correct to the nearest centimetre. This gives us the bounds for PQ and PR:

PQ = 4 \text{ cm} \implies 3.5 \le PQ < 4.5

PR = 2 \text{ cm} \implies 1.5 \le PR < 2.5

To find the minimum possible length of QR, we must use the minimum value of PQ (

PQ_{min}

) and the maximum value of PR (

PR_{max}

).

QR_{min} = \sqrt{PQ_{min}^2 - PR_{max}^2}

Substitute the boundary values:

QR_{min} = \sqrt{3.5^2 - 2.5^2}

Using the difference of two squares,

a^2 - b^2 = (a-b)(a+b)

:

QR_{min} = \sqrt{(3.5 - 2.5)(3.5 + 2.5)}

QR_{min} = \sqrt{(1)(6)} = \sqrt{6}

Question 52

1 mark

The graph shows the variation with time of the height through which a crane lifts a mass of 20 kg.

What is the power output of the crane when the mass is at a height of 10 m?

(gravitational field strength =

10 \text{Nkg}^{-1}

, the effects of air resistance and friction are negligible)

A.0.1 W
B.10 W
C.40 W
D.100 W
E.400 W
F.4000 W

Answer: D

Worked solution

The power output of the crane is the rate at which it does work. Since the effects of air resistance and friction are negligible, the work done is equal to the gain in gravitational potential energy.

\text{Power} = \frac{\text{Work Done}}{\text{Time}} = \frac{\Delta E_p}{\Delta t}

The change in potential energy is

\Delta E_p = mg\Delta h

. Thus,

\text{Power} = \frac{mg\Delta h}{\Delta t} = mg \left(\frac{\Delta h}{\Delta t}\right) = mgv

The velocity,

v

, is the gradient of the height-time graph. At the point where the height is

10 \text{ m}

, the graph is linear. We can calculate the gradient for this section:

v = \text{gradient} = \frac{15\text{ m} - 5\text{ m}}{30\text{ s} - 10\text{ s}} = \frac{10\text{ m}}{20\text{ s}} = 0.5 \text{ m/s}

Now, we calculate the power with

m = 20 \text{ kg}

and

g = 10 \text{ Nkg}^{-1}

.

P = (20 \text{ kg}) \times (10 \text{ Nkg}^{-1}) \times (0.5 \text{ m/s}) = 100 \text{ W}

Question 53

1 mark

The angle x is measured in radians and is such that

0 \le x \le \pi

.

The total length of any intervals for which

-1 \le \tan x \le 1

and

\sin 2x \ge 0.5

is

A. $\frac{\pi}{12}$
B. $\frac{\pi}{6}$
C. $\frac{\pi}{4}$
D. $\frac{\pi}{3}$
E. $\frac{5\pi}{12}$
F. $\frac{\pi}{2}$
G. $\frac{5\pi}{6}$

Answer: B

Worked solution

We must find the total length of the interval(s) where both inequalities are met for

x \in [0, \pi]

. First inequality:

-1 \le \tan x \le 1

. This is satisfied in the intervals:

0 \le x \le \frac{\pi}{4} \quad \text{and} \quad \frac{3\pi}{4} \le x \le \pi

Second inequality:

\sin(2x) \ge 0.5

. This is satisfied when:

\frac{\pi}{6} \le 2x \le \frac{5\pi}{6}

Dividing by 2 gives the interval for

x

:

\frac{\pi}{12} \le x \le \frac{5\pi}{12}

Intersection: We find the overlap between the solution sets. The only common region is:

\frac{\pi}{12} \le x \le \frac{\pi}{4}

Total length: The length of this interval is calculated as:

\text{Length} = \frac{\pi}{4} - \frac{\pi}{12} = \frac{3\pi - \pi}{12} = \frac{2\pi}{12} = \frac{\pi}{6}

Question 54

1 mark

A train consists of a powered engine travelling horizontally pulling two unpowered carriages.

[Diagram showing: carriage 2 --T-- carriage 1 -- powered engine -> 15 000 N]

The engine has a mass of 20 000 kg, and each carriage has a mass of 5000 kg. When the engine accelerates from rest it develops a thrust (driving force) of 15 000 N as shown.

Ignoring resistive forces, what is the tension (pulling force) T in the light and inextensible coupling between carriage 1 and carriage 2?

A.2500 N
B.3750 N
C.5000 N
D.7500 N
E.15 000 N

Answer: A

Worked solution

First, we find the acceleration of the entire system.
ewline The total mass of the train is:

m_{\text{total}} = 20000\,\text{kg} + 5000\,\text{kg} + 5000\,\text{kg} = 30000\,\text{kg}

Using Newton's second law,

F=ma

, with the given thrust:

a = \frac{F}{m_{\text{total}}} = \frac{15000\,\text{N}}{30000\,\text{kg}} = 0.5\,\text{m/s}^2

The tension

T

in the coupling between carriage 1 and carriage 2 is the net force acting on carriage 2. Applying Newton's second law to carriage 2:

T = m_{\text{carriage 2}} \times a \ = 5000\,\text{kg} \times 0.5\,\text{m/s}^2 \ = 2500\,\text{N}