VCAA Physics How do things move without contact?

15 sample questions with marking guides and sample answers

Q17
2020
QCAA
Paper 1
1 mark
Q17
1 mark

The definition of magnetic field is

A

a region of space through which the total magnetic flux is measured.

B

a region of space surrounding a body in which another body experiences a force of attraction.

C

a region of space around an electrically charged particle or object within which a force would be exerted on other electrically charged particles or objects.

D

a region of space near a magnet, electric current or moving electrically charged particle in which a magnetic force acts on any other magnet, electric current or moving electrically charged particle.

Reveal Answer
A

a region of space through which the total magnetic flux is measured.

This describes the domain for calculating magnetic flux, which is the integral of the magnetic field over an area, rather than defining the field itself.

B

a region of space surrounding a body in which another body experiences a force of attraction.

This is a generic definition that could apply to gravitational or electric fields; it fails to specify that the force is magnetic or that it acts on magnets and moving charges.

C

a region of space around an electrically charged particle or object within which a force would be exerted on other electrically charged particles or objects.

This is the definition of an electric field, which exerts forces on stationary or moving electric charges, whereas magnetic fields specifically affect moving charges or magnetic materials.

D

a region of space near a magnet, electric current or moving electrically charged particle in which a magnetic force acts on any other magnet, electric current or moving electrically charged particle.

Correct Answer

This accurately defines a magnetic field by identifying its sources (magnets, currents, moving charges) and its effect (exerting magnetic force on similar entities).

Q33
2023
NESA
9 marks
Q33
9 marks

Consider the following statement.

"The interaction of subatomic particles with fields, as well as with other types of particles and matter, has increased our understanding of processes that occur in the physical world and of the properties of the subatomic particles themselves."

Justify this statement with reference to observations that have been made and experiments that scientists have carried out.

Reveal Answer
Marking Criteria
DescriptorMarks

Provides a reasoned detailed justification for the statement with explanation referencing at least TWO observations and at least TWO experiments

9

Provides a justification for the statement with explanation referencing at least TWO observations and at least TWO experiments

8

The student response meets all criteria of the 6-mark band, and additionally meets the majority of criteria in the 8-mark band.

7

Provides a justification for the statement with an explanation of an observation/experiment of particle–field interactions and an observation/experiment of particle–particle interactions

6

The student response meets all criteria of the 4-mark band, and additionally meets the majority of criteria in the 6-mark band.

5

Provides details of TWO experiments or observations and how they relate to the statement

4

The student response meets all criteria of the 2-mark band, and additionally meets the majority of criteria in the 4-mark band.

3
  • Relates the statement to an experiment or observation
    OR
  • Describes relevant experiments or observations
2

Provides some relevant information

1

None of the above

0
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.

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.

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
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
Q7
2022
QCAA
Paper 1
1 mark
Q7
1 mark

Which change would produce the greatest increase in magnetic field strength inside a current-carrying solenoid?

A

decreasing the thickness of the wire

B

increasing the length of the solenoid

C

adding more turns of wire to the solenoid

D

using an alternating current instead of a direct current

Reveal Answer
A

decreasing the thickness of the wire

Decreasing wire thickness increases electrical resistance, which would reduce the current II (for a fixed voltage) and consequently decrease the magnetic field strength.

B

increasing the length of the solenoid

The magnetic field strength B=μ0NLIB = \mu_0 \frac{N}{L} I is inversely proportional to the length LL; increasing the length while keeping the number of turns constant decreases the turn density, weakening the field.

C

adding more turns of wire to the solenoid

Correct Answer

The magnetic field strength inside a solenoid is directly proportional to the number of turns (NN); increasing the number of turns increases the turn density nn, which increases the field strength according to B=μ0nIB = \mu_0 n I.

D

using an alternating current instead of a direct current

Alternating current produces a magnetic field that fluctuates in magnitude and direction, and inductive reactance often reduces the current compared to DC, which would not increase the field strength.

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
Q10
2024
QCAA
Paper 1
1 mark
Q10
1 mark

An experiment was conducted to determine the force experienced by an 85 cm wire with a 2.4 A current flowing through it in an external magnetic field. It was rotated through varying angles within the magnetic field such that data analysis identified the relationship F=0.0306sinθF = 0.0306 \sin \theta.

What is the order of magnitude of the strength of the external magnetic field?

A

10410^{-4} T

B

10210^{-2} T

C

10210^{2} T

D

10410^{4} T

Reveal Answer
A

10410^{-4} T

This value is too small. The calculated magnetic field strength is approximately 0.015 T0.015 \text{ T}, which is two orders of magnitude larger than 104 T10^{-4} \text{ T}.

B

10210^{-2} T

Correct Answer

Comparing the experimental relationship F=0.0306sinθF = 0.0306 \sin \theta to the theoretical formula F=ILBsinθF = ILB \sin \theta, we see that ILB=0.0306ILB = 0.0306. Solving for BB gives B=0.03062.4×0.850.015 TB = \frac{0.0306}{2.4 \times 0.85} \approx 0.015 \text{ T}, which is on the order of 102 T10^{-2} \text{ T}.

C

10210^{2} T

This value is significantly larger than the actual magnetic field. The calculation yields B0.015 TB \approx 0.015 \text{ T}, whereas 102 T10^2 \text{ T} represents an extremely strong magnetic field not supported by the data.

D

10410^{4} T

This value is far too large. The calculated field strength is approximately 1.5×102 T1.5 \times 10^{-2} \text{ T}, which is six orders of magnitude smaller than 104 T10^4 \text{ T}.

Q19
2023
QCAA
Paper 1
1 mark
Q19
1 mark

Calculate the electric field strength experienced at a distance of 2.8×10112.8\times10^{-11} m from the centre of a helium nucleus.

A

1.0×102 N C11.0\times10^2\text{ N C}^{-1}

B

2.0×102 N C12.0\times10^2\text{ N C}^{-1}

C

3.7×1012 N C13.7\times10^{12}\text{ N C}^{-1}

D

7.3×1012 N C17.3\times10^{12}\text{ N C}^{-1}

Reveal Answer
A

1.0×102 N C11.0\times10^2\text{ N C}^{-1}

This is the value for the electric potential (V=kQrV = \frac{kQ}{r}), not the electric field strength (E=kQr2E = \frac{kQ}{r^2}).

B

2.0×102 N C12.0\times10^2\text{ N C}^{-1}

This value represents the electric potential calculated incorrectly by using the mass number (4) instead of the atomic number (2) for the charge.

C

3.7×1012 N C13.7\times10^{12}\text{ N C}^{-1}

Correct Answer

Using the formula E=kQr2E = \frac{kQ}{r^2} with the charge of a helium nucleus (Q=2e=3.2×1019Q = 2e = 3.2 \times 10^{-19} C), the result is 3.7×1012 N C13.7 \times 10^{12} \text{ N C}^{-1}.

D

7.3×1012 N C17.3\times10^{12}\text{ N C}^{-1}

This result comes from incorrectly using the mass number (4) instead of the atomic number (2) to calculate the total charge (Q=4eQ = 4e).

Q5
2021
QCAA
Paper 2
4 marks
Q5
4 marks

An alpha particle with a charge of +3.2×1019+3.2 \times 10^{-19} C moves through an electric field, accelerating from rest through a potential difference of 240 V.

Determine the velocity of the particle at the end of its acceleration, expressing your answer in scientific notation. (m/s to 2 significant figures)

Reveal Answer

The change in potential energy of an electric charge moving through an electric field is equivalent to the work done on the charge.
V=ΔUqV = \frac{\Delta U}{q}
ΔU=Vq\Delta U = Vq
=240×3.2×1019= 240 \times 3.2 \times 10^{-19}
=7.68×1017J=W= 7.68 \times 10^{-17} J = W

The work done on an object is equal to the change in kinetic energy.
Ek=12mv2E_k = \frac{1}{2}mv^2
7.68×1017=12×6.64×1027×v27.68 \times 10^{-17} = \frac{1}{2} \times 6.64 \times 10^{-27} \times v^2
v2=7.68×101712×6.64×1027v^2 = \frac{7.68 \times 10^{-17}}{\frac{1}{2} \times 6.64 \times 10^{-27}}
v=7.68×101712×6.64×1027v = \sqrt{\frac{7.68 \times 10^{-17}}{\frac{1}{2} \times 6.64 \times 10^{-27}}}
Velocity = 1.5×105 m s11.5 \times 10^5 \text{ m s}^{-1} (to 2 significant figures)

Marking Criteria
DescriptorMarks

Recognises the scenario relates to work done on a moving charge in an electric field

1

Identifies that work done on the charge equates to its kinetic energy

1

Provides appropriate mathematical reasoning

1

Determines the velocity

1
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
Q17
2021
QCAA
Paper 1
1 mark
Q17
1 mark

Electrical potential energy is the

A

intensity of an electric field at a particular location.

B

difference in potential that tends to give rise to an electric current.

C

capacity of electric charge carriers to do work due to their position in an electric circuit.

D

work done on an electron in accelerating it through an electrical potential difference of one volt.

Reveal Answer
A

intensity of an electric field at a particular location.

This describes electric field strength (EE), which is the force exerted per unit charge at a specific location, not the energy.

B

difference in potential that tends to give rise to an electric current.

This describes electric potential difference (voltage), which is the difference in electric potential energy per unit charge between two points.

C

capacity of electric charge carriers to do work due to their position in an electric circuit.

Correct Answer

Electrical potential energy is the stored energy a charge possesses due to its position in an electric field, representing the capacity to do work.

D

work done on an electron in accelerating it through an electrical potential difference of one volt.

This is the specific definition of an electron-volt (eV), a unit of energy, rather than the general definition of electrical potential energy.

Q14
2023
NESA
1 mark
Q14
1 mark

Planet X has a mass 4 times that of Earth and a radius 3 times that of Earth. The escape velocity at the surface of Earth is 11.2 km s1^{-1}.

What is the escape velocity at the surface of planet X?

A

8.40 km s1^{-1}

B

9.70 km s1^{-1}

C

12.9 km s1^{-1}

D

14.9 km s1^{-1}

Reveal Answer
A

8.40 km s1^{-1}

Incorrect. This value is obtained by multiplying Earth's escape velocity by 3/43/4, incorrectly assuming escape velocity is proportional to R/MR/M.

B

9.70 km s1^{-1}

Incorrect. This value is obtained by multiplying Earth's escape velocity by 3/4\sqrt{3/4}, incorrectly assuming escape velocity is proportional to R/M\sqrt{R/M}.

C

12.9 km s1^{-1}

Correct Answer

Correct. Escape velocity is given by v=2GM/Rv = \sqrt{2GM/R}, meaning it is proportional to M/R\sqrt{M/R}. For Planet X, vX=vE4/312.9v_X = v_E \sqrt{4/3} \approx 12.9 km s1^{-1}.

D

14.9 km s1^{-1}

Incorrect. This value is obtained by multiplying Earth's escape velocity by 4/34/3, incorrectly assuming escape velocity is proportional to M/RM/R instead of M/R\sqrt{M/R}.

Q27
2023
QCAA
Paper 1
5 marks
Q27
5 marks

A satellite orbits a planet of mass 6.42×10236.42\times 10^{23} kg at a height of 5000 km from the surface. The planet has a diameter of 6780 km.

Determine the speed required for the satellite to maintain its orbit. Show your working.
Speed = _____ ms1ms^{-1} (two significant figures)

Reveal Answer

T2r3=4π2GM\frac{T^2}{r^3} = \frac{4\pi^2}{GM}

T2(3390×103+5000×103)3=4π26.67×1011×6.42×1023\frac{T^2}{(3390 \times 10^3 + 5000 \times 10^3)^3} = \frac{4\pi^2}{6.67 \times 10^{-11} \times 6.42 \times 10^{23}}

T=4π2×5.91×10206.67×1011×6.42×1023T = \sqrt{\frac{4\pi^2 \times 5.91 \times 10^{20}}{6.67 \times 10^{-11} \times 6.42 \times 10^{23}}}
=2.33×104 s= 2.33 \times 10^4 \text{ s}

v=2πrTv = \frac{2\pi r}{T}
=2π×8.39×1062.33×104= \frac{2\pi \times 8.39 \times 10^6}{2.33 \times 10^4}
=2.3×103 m s1= 2.3 \times 10^3 \text{ m s}^{-1}

Speed =2.3×103 m s1= 2.3 \times 10^3 \text{ m s}^{-1} (to two significant figures)

Marking Criteria
DescriptorMarks

Recognises the scenario relates to orbital mechanics

1

Recognises the scenario relates to circular motion

1

Provides appropriate mathematical reasoning

1

Demonstrates correct substitution

1

Calculates the speed

1

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