SCSA Physics Science Understanding

15 sample questions with marking guides and sample answers

Q12
2022
QCAA
Paper 1
1 mark
Q12
1 mark

An object experiencing uniform circular motion in a horizontal plane travels at an average speed of 8.0 m s18.0 \text{ m s}^{-1}.
Calculate the radius of the object’s path if it takes 0.3 s to complete a full rotation.

A

3.8×1013.8 \times 10^{-1} m

B

2.6×1002.6 \times 10^0 m

C

1.5×1011.5 \times 10^1 m

D

1.7×1021.7 \times 10^2 m

Reveal Answer
A

3.8×1013.8 \times 10^{-1} m

Correct Answer

This is the correct answer. Using the formula for speed in uniform circular motion, v=2πrTv = \frac{2\pi r}{T}, we can rearrange for radius: r=vT2πr = \frac{vT}{2\pi}. Substituting the values gives r=8.0×0.32π0.38 mr = \frac{8.0 \times 0.3}{2\pi} \approx 0.38 \text{ m}.

B

2.6×1002.6 \times 10^0 m

This option is incorrect. It is close to the value of the circumference (C=vT=2.4 mC = vT = 2.4 \text{ m}) or the result of dividing speed by π\pi, rather than solving for the radius using 2π2\pi.

C

1.5×1011.5 \times 10^1 m

This option is incorrect and results from a calculation error or misapplication of the circular motion variables.

D

1.7×1021.7 \times 10^2 m

This option is incorrect. It results from incorrectly rearranging the formula as r=2πvTr = \frac{2\pi v}{T} instead of dividing by 2π2\pi.

Q4
2023
QCAA
Paper 2
3 marks
Q4
3 marks

Two objects on different planets experience different accelerations due to gravity.

ObjectMass (kg)Acceleration due to gravity (m s2^{-2})
A791.6
B323.7

Determine which object has the greatest force acting on it. Show your working.

Reveal Answer

Force on object A = mg=79×1.6=126.4 N130 Nmg = 79 \times 1.6 = 126.4 \text{ N} \approx 130 \text{ N} down

Force on object B = mg=32×3.7=118 N120 Nmg = 32 \times 3.7 = 118 \text{ N} \approx 120 \text{ N} down

Object A experiences the greatest force.

Marking Criteria
DescriptorMarks

Recognises the scenario relates to relationship between the force due to gravity and mass

1

Provides appropriate mathematical reasoning

1

Identifies the object experiencing the greatest force acting on it

1
Q19
2021
QCAA
Paper 1
1 mark
Q19
1 mark

A spaceship with a velocity of 9.0×1079.0 \times 10^7 m s1^{-1} is measured to be 125 m in length by an observer at rest.
Calculate the length of the spaceship as measured by somebody on board the spaceship.

A

119 m

B

131 m

C

137 m

D

178 m

Reveal Answer
A

119 m

This calculation incorrectly treats the given 125 m as the proper length (L0L_0) and solves for the contracted length. Since the observer on the spaceship is at rest relative to it, they measure the proper length, which must be longer than the contracted length measured by the outside observer.

B

131 m

Correct Answer

The observer on board measures the proper length (L0L_0). Using the length contraction formula L=L01v2c2L = L_0 \sqrt{1 - \frac{v^2}{c^2}}, with L=125L=125 m and v=0.3cv=0.3c, we rearrange to find L0=12510.09131L_0 = \frac{125}{\sqrt{1 - 0.09}} \approx 131 m.

C

137 m

This answer results from omitting the square root in the Lorentz factor calculation, dividing 125 by (10.32)(1 - 0.3^2) instead of 10.32\sqrt{1 - 0.3^2}.

D

178 m

This value is incorrect and does not align with the standard length contraction formula. It likely results from a calculation error or misapplication of the Lorentz factor γ\gamma.

Q9
2021
SCSA
5 marks
Q9
5 marks

A space station is shaped like a huge hollow doughnut that is rotating uniformly. The outer radius is 4.60 × 10² m. What is the period of rotation of the station if a person standing on the outer wall inside the station experiences the same weight force she would experience on Earth?

[Copyrighted image]

Reveal Answer

The centripetal force is supplied by the reaction force, so mv2/r=Rmv^2/r = R.

The reaction force equals mgmg, giving mv2/r=mgmv^2/r = mg.

Rearranging the formula to calculate velocity gives v=rg=4.60×102×9.80=67.1 m s1v = \sqrt{rg} = \sqrt{4.60 \times 10^2 \times 9.80} = 67.1 \text{ m s}^{-1}.

The period is circumference over time, T=2πr/vT = 2\pi r/v.

Calculating the period gives T=43.0 sT = 43.0 \text{ s}.

Marking Criteria
DescriptorMarks

States that centripetal force is supplied by the reaction force (mv2/r=Rmv^2/r = R)

1

Equates reaction force to weight (mv2/r=mgmv^2/r = mg)

1

Correctly rearranges formula and calculates velocity (v=rg=4.60×102×9.80=67.1 m s1v = \sqrt{rg} = \sqrt{4.60 \times 10^2 \times 9.80} = 67.1 \text{ m s}^{-1})

1

States that period is circumference over time (T=2πr/vT = 2\pi r/v)

1

Correctly calculates period (T=43.0 sT = 43.0 \text{ s})

1
Q7
2022
QCAA
Paper 2
3 marks
Q7
3 marks

Two asteroids experience a gravitational force of 3.3×1033.3 \times 10^3 N between them. Their masses are 2.7×10172.7 \times 10^{17} kg and 6.1×10156.1 \times 10^{15} kg.

Calculate the distance between the two asteroids. Show your working. (m to two significant figures)

Reveal Answer

F=GMmr2F = \frac{GMm}{r^2}
3.3×103=6.67×1011×2.7×1017×6.1×1015r23.3 \times 10^3 = \frac{6.67 \times 10^{-11} \times 2.7 \times 10^{17} \times 6.1 \times 10^{15}}{r^2}
r=6.67×1011×2.7×1017×6.1×10153.3×103r = \sqrt{\frac{6.67 \times 10^{-11} \times 2.7 \times 10^{17} \times 6.1 \times 10^{15}}{3.3 \times 10^3}}
=5.8×109 m= 5.8 \times 10^9\ m

Distance between asteroids = 5.8×1095.8 \times 10^9 m (to two significant figures)

Marking Criteria
DescriptorMarks

Recognises the scenario relates to Newton’s Law of Universal Gravitation

1

Provides appropriate mathematical reasoning

1

Calculates the distance between the asteroids

1
Q2
2024
QCAA
Paper 1
1 mark
Q2
1 mark

Which property of light is described by the postulates of special relativity?

A

The energy of light is greater when the frequency of the photons decreases.

B

The wavelength of light decreases as the velocity of the source increases.

C

The velocity of light remains constant in all inertial frames of reference.

D

The frequency of light changes depending on media.

Reveal Answer
A

The energy of light is greater when the frequency of the photons decreases.

This statement is factually incorrect and unrelated to special relativity. According to the quantum relation E=hfE=hf, energy is directly proportional to frequency, so energy decreases as frequency decreases.

B

The wavelength of light decreases as the velocity of the source increases.

This describes the Doppler effect rather than a fundamental postulate. While relativity influences the Doppler shift, the postulates specifically define the behavior of the speed of light, not the change in wavelength due to source motion.

C

The velocity of light remains constant in all inertial frames of reference.

Correct Answer

This is the second postulate of special relativity. It states that the speed of light in a vacuum (cc) is constant and independent of the motion of the light source or the observer in all inertial frames.

D

The frequency of light changes depending on media.

This is incorrect regarding wave mechanics; frequency remains constant when light changes media, while speed and wavelength change. Furthermore, this is a concept of optics, not a postulate of special relativity.

Q22
2023
QCAA
Paper 1
3 marks
Q22
3 marks

Particles move at a rate of 1.3×1061.3\times10^6 times per second around a circular particle accelerator with a radius of 35 m.

Calculate the average speed of the particles. Show your working.

Average speed = ______ ms1ms^{-1} (two significant figures)

Reveal Answer

C=2πr=2×π×35219.91 mC = 2\pi r = 2 \times \pi \times 35 \approx 219.91 \text{ m}

f=1Tf = \frac{1}{T}

v=2πrT=219.91×1.3×106=2.86×108 m s1\therefore v = \frac{2\pi r}{T} = 219.91 \times 1.3 \times 10^6 = 2.86 \times 10^8 \text{ m s}^{-1}

Average speed =2.9×108 m s1= 2.9 \times 10^8 \text{ m s}^{-1} (to two significant figures)

Marking Criteria
DescriptorMarks

Recognises the scenario relates to average speed of objects in uniform circular motion

1

Provides appropriate mathematical reasoning

1

Calculates the average speed of the particles

1
Q4
2023
QCAA
Paper 1
1 mark
Q4
1 mark

Kepler’s third law

A

describes the elliptical orbit of planets.

B

combines Newton’s first law of motion with uniform circular motion.

C

equates the area of the arc sweep of a planet to the time taken to complete it.

D

describes the relationship between uniform circular motion and the Law of Universal Gravitation.

Reveal Answer
A

describes the elliptical orbit of planets.

This describes Kepler's First Law, also known as the Law of Ellipses, which states that planets move in elliptical orbits with the Sun at one focus.

B

combines Newton’s first law of motion with uniform circular motion.

Kepler's laws are not derived from combining Newton's first law with uniform circular motion; rather, the third law relates orbital period to distance.

C

equates the area of the arc sweep of a planet to the time taken to complete it.

This describes Kepler's Second Law, or the Law of Equal Areas, which states that a line segment joining a planet and the Sun sweeps out equal areas during equal intervals of time.

D

describes the relationship between uniform circular motion and the Law of Universal Gravitation.

Correct Answer

Kepler's Third Law (T2r3T^2 \propto r^3) is physically derived by equating the centripetal force in uniform circular motion to the gravitational force defined by Newton's Law of Universal Gravitation.

Q11
2020
VCAA
4 marks
Q11

An astronaut has left Earth and is travelling on a spaceship at 0.800c (γ = 1.67) directly towards the star known as Sirius, which is located 8.61 light-years away from Earth, as measured by observers on Earth.

Q11a
2 marks

How long will the trip take according to a clock that the astronaut is carrying on his spaceship? Show your working.

Reveal Answer

The distance of 8.61 light-years is the proper length in Earth's frame of reference. The proper time in the Earth's frame of reference is:
t=dv=8.610.8t = \frac{d}{v} = \frac{8.61}{0.8}
t=10.76 yrt = 10.76 \text{ yr}
This time is the dilated time in the astronaut's frame of reference. The proper time as measured by the astronaut is:
t=t0γt = t_0 \gamma
10.76=t0×1.6710.76 = t_0 \times 1.67
t0=6.44 yearst_0 = 6.44 \text{ years}

Marking Criteria
DescriptorMarks

Calculates the time in Earth's frame of reference (10.76 yr10.76 \text{ yr}) OR calculates the contracted length in the astronaut's frame of reference (5.16 ly5.16 \text{ ly})

1

Calculates the correct proper time of 6.44 years6.44 \text{ years}

1
Q11b
2 marks

Is the trip time measured by the astronaut in part a. a proper time? Explain your reasoning.

Reveal Answer

The time measured by the astronaut will be proper time because the clock is stationary in the astronaut's frame of reference.

Marking Criteria
DescriptorMarks

Identifies that the time measured by the astronaut is proper time

1

Explains that this is because the clock is stationary in the astronaut's frame of reference

1
Q2
2024
VCAA
1 mark
Q2
1 mark

A space-based observatory (SBO) of mass MM has a circular orbital radius RR around Earth. Modifications to the SBO have doubled its mass, but its orbital speed is kept constant.

Which one of the following is closest to the orbital radius of the SBO after the modifications have been made?

A

R4\frac{R}{4}

B

RR

C

2R2R

D

4R4R

Reveal Answer
A

R4\frac{R}{4}

This assumes the orbital radius is inversely proportional to the square of the satellite's mass. However, orbital speed and radius are completely independent of the satellite's mass.

B

RR

Correct Answer

The orbital speed v=GMERv = \sqrt{\frac{G M_E}{R}} depends only on Earth's mass and the orbital radius, not the satellite's mass. Since the speed is kept constant, the orbital radius must remain RR.

C

2R2R

This incorrectly assumes the orbital radius is directly proportional to the satellite's mass. The mass of the orbiting object cancels out when equating gravitational and centripetal forces.

D

4R4R

This incorrectly assumes the orbital radius is proportional to the square of the satellite's mass. A satellite's mass has no effect on its orbital radius for a given constant speed.

Q18
2025
NESA
1 mark
Q18
1 mark

The escape velocity from the surface of a planet, which has no atmosphere, is vv. A mass is launched at 4545^\circ to the planet's surface at vv.

What will be the subsequent motion of the mass?

A

A circular orbit around the planet

B

An elliptical orbit around the planet

C

A parabolic trajectory, returning to land with velocity vv

D

A trajectory reaching zero velocity at an infinite distance

Reveal Answer
A

A circular orbit around the planet

A circular orbit requires a negative total energy and a launch parallel to the surface at a specific orbital velocity, which is less than the escape velocity.

B

An elliptical orbit around the planet

An elliptical orbit requires the total energy of the system to be negative, meaning the launch velocity must be strictly less than the escape velocity.

C

A parabolic trajectory, returning to land with velocity vv

While the trajectory is indeed a parabola, an object launched at or above escape velocity will overcome the planet's gravity and never return to land.

D

A trajectory reaching zero velocity at an infinite distance

Correct Answer

At escape velocity, the total energy (kinetic plus gravitational potential) of the mass is exactly zero. Regardless of the outward launch angle, it will escape the planet's gravitational field and reach zero velocity at an infinite distance.

Q24
2020
QCAA
Paper 1
4 marks
Q24
4 marks

A spaceship travelled from Planet A to Planet B at a speed of 0.90c0.90c. An observer that was stationary relative to both planets measured the time taken for the trip to be 4.0 years.
Calculate the time taken for the trip as measured by an observer on the spaceship. (years to 1 decimal place)

Reveal Answer

t=t01v2c2t = \frac{t_0}{\sqrt{1-\frac{v^2}{c^2}}}

4.0=t01(0.90c)2c24.0 = \frac{t_0}{\sqrt{1-\frac{(0.90c)^2}{c^2}}}

t0=1.7t_0 = 1.7 years

Time = 1.7 years (to 1 decimal place)

Marking Criteria
DescriptorMarks

Indicates an understanding of the physical scenario in relation to time dilation (or other relevant physical concept/s).

1

Indicates an understanding that the time provided in the question represents relativistic (dilated) time. (If proper time and relativistic time are confused, a maximum of 2 marks can be awarded overall.)

1

Provides pertinent mathematical operation/s correctly performed.

1

Determines the time correctly (accept 1.7 years to 1.8 years inclusive).

1
Q15
2024
QCAA
Paper 1
1 mark
Q15
1 mark

In which direction does the centripetal force act?

A

towards the centre of motion

B

away from the centre of motion

C

opposite to the object's direction of motion

D

tangentially to the object's direction of motion

Reveal Answer
A

towards the centre of motion

Correct Answer

The term "centripetal" means "center-seeking," and this force acts perpendicular to the velocity vector, directed radially inward, to constantly change the object's direction.

B

away from the centre of motion

This describes the direction of the apparent "centrifugal" force (a pseudo-force); the actual centripetal force must pull inward to keep the object on a curved path.

C

opposite to the object's direction of motion

A force acting opposite to the direction of motion acts as a braking force that slows the object down, rather than causing the perpendicular acceleration required for circular motion.

D

tangentially to the object's direction of motion

A tangential force acts parallel to the velocity and changes the object's speed, whereas centripetal force acts perpendicular to the velocity to change the direction.

Q24
2025
NESA
3 marks
Q24
3 marks

Two satellites, AA and BB, are in stable circular orbits around the Earth. The radius of satellite AA's orbit is three times that of satellite BB's orbit. Both satellites have the same kinetic energy.

Show that the mass of AA is three times the mass of BB.

Reveal Answer

FC=FGmv2r=GMmr2mv2=GMmr12mv2=GMm2rGMmA2rA=GMmB2rB\begin{align*} F_C &= F_G\\ \frac{mv^2}{r} &= \frac{GMm}{r^2}\\ mv^2 &= \frac{GMm}{r}\\ \frac{1}{2}mv^2 &= \frac{GMm}{2r}\\ \therefore \frac{G M m_A}{2 r_A} &= \frac{G M m_B}{2 r_B} \end{align*}

Substitute rA=3rBr_A = 3r_B:

3rBrB=mAmB=3\frac{3r_B}{r_B} = \frac{m_A}{m_B} = 3

mA=3mB\therefore m_A = 3m_B

Marking Criteria
DescriptorMarks

Shows all relevant steps to determine the mass ratio

3

Makes progress towards determining mass ratio

2

Provides some relevant information

1

None of the above

0
Q18
2022
VCAA
1 mark
Q18
1 mark

Which one of the following is an example of an inertial frame of reference?

A

a bus travelling at constant velocity

B

an express train that is accelerating

C

a car turning a corner at a constant speed

D

a roller-coaster speeding up while heading down a slope

Reveal Answer
A

a bus travelling at constant velocity

Correct Answer

This is correct because an inertial frame of reference is one that is not accelerating. A bus moving at a constant velocity has zero acceleration, making it an inertial frame where Newton's laws of motion hold true.

B

an express train that is accelerating

This is incorrect because an accelerating object represents a non-inertial frame of reference. By definition, an inertial frame must have zero acceleration.

C

a car turning a corner at a constant speed

This is incorrect because turning a corner involves a change in direction, which means the car is experiencing centripetal acceleration. Even at a constant speed, a change in direction makes it a non-inertial frame of reference.

D

a roller-coaster speeding up while heading down a slope

This is incorrect because the roller-coaster is speeding up, meaning its velocity is changing and it is accelerating. Any accelerating environment is a non-inertial frame of reference.

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