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VCE Physics — How do physicists explain motion in two dimensions? Questions & 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 $8.0 \text{ m s}^{-1}$ .
Calculate the radius of the object’s path if it takes 0.3 s to complete a full rotation.

A

$3.8 \times 10^{-1}$ m

B

$2.6 \times 10^0$ m

C

$1.5 \times 10^1$ m

D

$1.7 \times 10^2$ m

Reveal Answer

A

$3.8 \times 10^{-1}$ m

Correct Answer

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

B

$2.6 \times 10^0$ m

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

C

$1.5 \times 10^1$ m

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

D

$1.7 \times 10^2$ m

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

Q5

2021

QCAA

Paper 2

4 marks

Q5

4 marks

An alpha particle with a charge of $+3.2 \times 10^{-19}$ 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.
$V = \frac{\Delta U}{q}$
$\Delta U = Vq$
$= 240 \times 3.2 \times 10^{-19}$
$= 7.68 \times 10^{-17} J = W$

The work done on an object is equal to the change in kinetic energy.
$E_k = \frac{1}{2}mv^2$
$7.68 \times 10^{-17} = \frac{1}{2} \times 6.64 \times 10^{-27} \times v^2$
$v^2 = \frac{7.68 \times 10^{-17}}{\frac{1}{2} \times 6.64 \times 10^{-27}}$
$v = \sqrt{\frac{7.68 \times 10^{-17}}{\frac{1}{2} \times 6.64 \times 10^{-27}}}$
Velocity = $1.5 \times 10^5 \text{ m s}^{-1}$ (to 2 significant figures)

Marking Criteria

Descriptor	Marks
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

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?

Reveal Answer

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

The reaction force equals $mg$ , giving $mv^2/r = mg$ .

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

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

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

Marking Criteria

Descriptor	Marks
States that centripetal force is supplied by the reaction force ( $mv^2/r = R$ )	1
Equates reaction force to weight ( $mv^2/r = mg$ )	1
Correctly rearranges formula and calculates velocity ( $v = \sqrt{rg} = \sqrt{4.60 \times 10^2 \times 9.80} = 67.1 \text{ m s}^{-1}$ )	1
States that period is circumference over time ( $T = 2\pi r/v$ )	1
Correctly calculates period ( $T = 43.0 \text{ s}$ )	1

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 s $^{-1}$ .

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

A

8.40 km s $^{-1}$

B

9.70 km s $^{-1}$

C

12.9 km s $^{-1}$

D

14.9 km s $^{-1}$

Reveal Answer

A

8.40 km s $^{-1}$

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

B

9.70 km s $^{-1}$

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

C

12.9 km s $^{-1}$

Correct Answer

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

D

14.9 km s $^{-1}$

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

Q4

2024

VCAA

1 mark

Q4

1 mark

Two satellites in circular orbits around Earth have the same mass but are travelling at different speeds.
This means that

A

they orbit Earth at different radii.

B

they will have the same kinetic energy.

C

they orbit Earth with the same period.

D

the gravitational force exerted on them by Earth will be the same.

Reveal Answer

A

they orbit Earth at different radii.

Correct Answer

The orbital speed of a satellite is given by $v = \sqrt{\frac{GM}{r}}$ . Since their speeds are different, their orbital radii must also be different.

B

they will have the same kinetic energy.

Kinetic energy is calculated as $KE = \frac{1}{2}mv^2$ . Since their speeds are different, their kinetic energies cannot be the same.

C

they orbit Earth with the same period.

According to Kepler's Third Law ( $T^2 \propto r^3$ ), satellites at different radii will have different orbital periods.

D

the gravitational force exerted on them by Earth will be the same.

The gravitational force depends on the orbital radius ( $F = \frac{GMm}{r^2}$ ). Since their radii are different, the forces exerted on them will also be different.

Q20

2025

VCAA

1 mark

Q20

1 mark

Harriet and Tom were investigating how the speed, $v$ , of a falling object varied with the distance, $s$ , it had fallen.

They dropped a small steel ball, initially at rest, from the third floor of their school building. The speed of the ball was measured at six positions as it fell.

Air resistance can be ignored.

Which one of the following graphs of their data would be expected to result in a straight line through the origin?

A

$v$ versus $s$

B

$v$ versus $\sqrt{s}$

C

$v^2$ versus $\sqrt{s}$

D

$\sqrt{v}$ versus $s$

Reveal Answer

A

$v$ versus $s$

The kinematic equation for an object falling from rest is $v^2 = 2gs$ , meaning $v$ is proportional to $\sqrt{s}$ . A graph of $v$ versus $s$ would result in a curve, not a straight line.

B

$v$ versus $\sqrt{s}$

Correct Answer

Using the kinematic equation $v^2 = u^2 + 2as$ with an initial velocity of $u=0$ , we get $v = \sqrt{2g}\sqrt{s}$ . This shows that $v$ is directly proportional to $\sqrt{s}$ , which produces a straight line through the origin.

C

$v^2$ versus $\sqrt{s}$

Based on the equation $v^2 = 2gs$ , $v^2$ is directly proportional to $s$ , not $\sqrt{s}$ . Plotting $v^2$ versus $\sqrt{s}$ would result in a quadratic curve.

D

$\sqrt{v}$ versus $s$

Since $v$ is proportional to $s^{1/2}$ , $\sqrt{v}$ would be proportional to $s^{1/4}$ . Plotting $\sqrt{v}$ versus $s$ would not produce a straight line.

Q7

2024

QCAA

Paper 1

1 mark

Q7

1 mark

An object experiencing a gravitational force of 50.0 N moves down a frictionless incline of $40.0^\circ$ to the horizontal.
Calculate the net force acting on the object.

A

32.1 N

B

37.3 N

C

38.3 N

D

42.0 N

Reveal Answer

A

32.1 N

Correct Answer

On a frictionless incline, the net force is the component of gravity parallel to the slope, calculated as $F_{net} = F_g \sin(\theta) = 50.0 \, \text{N} \times \sin(40.0^\circ) \approx 32.1 \, \text{N}$ .

B

37.3 N

This incorrect value results from calculating the sine of the angle in radians ( $50.0 \sin(40 \text{ rad}) \approx 37.3 \, \text{N}$ ) rather than degrees.

C

38.3 N

This represents the perpendicular component of gravity ( $F_g \cos(40.0^\circ) \approx 38.3 \, \text{N}$ ), which is balanced by the normal force and does not contribute to the net force down the incline.

D

42.0 N

This value corresponds to $F_g \tan(40.0^\circ) \approx 42.0 \, \text{N}$ , which is an incorrect trigonometric relationship for finding the parallel force component.

Q3

2025

VCAA

5 marks

Q3

Anna, of mass $40.0 \text{ kg}$ , steps off a diving board at a height of $2.50 \text{ m}$ above the water.

Assume that air resistance is negligible.

Q3a

1 mark

Show that her kinetic energy just before hitting the water is $981 \text{ J}$ .

Reveal Answer

Using conservation of energy:

$\Delta KE = \Delta GPE$

$\Delta GPE = mg\Delta h$

$KE = 40.0 \times 9.81 \times 2.50$

$KE = 981 \text{ J}$

Marking Criteria

Descriptor	Marks
Shows correct substitution into the gravitational potential energy formula to calculate kinetic energy (e.g., $40.0 \times 9.81 \times 2.50$ )	1

Q3b

2 marks

Calculate her speed just before hitting the water.

Reveal Answer

$KE = \frac{1}{2}mv^2$

$981 = 0.5 \times 40.0 \times v^2$

$v = 7.00 \text{ m s}^{-1}$

Marking Criteria

Descriptor	Marks
Correctly substitutes values into the kinetic energy formula (e.g., $981 = 0.5 \times 40.0 \times v^2$ )	1
Calculates the correct final speed of $7.00 \text{ m s}^{-1}$	1

Q3c

2 marks

Determine the height Anna would have to drop from so that the speed reached just before hitting the water is double the answer to part b.

Reveal Answer

Required speed is now $14.0 \text{ m s}^{-1}$ .

By conservation of energy:

$\Delta KE = \Delta GPE$

$\frac{1}{2}mv^2 = mgh$

$0.5 \times 40.0 \times 14.0^2 = 40.0 \times 9.81 \times h$

$h = 10 \text{ m}$

Marking Criteria

Descriptor	Marks
Correctly substitutes double the speed from part b ( $14.0 \text{ m s}^{-1}$ ) into the conservation of energy equation (e.g., $0.5 \times 40.0 \times 14.0^2 = 40.0 \times 9.81 \times h$ )	1
Calculates the correct final height of $10 \text{ m}$	1

Q2

2024

VCAA

1 mark

Q2

1 mark

A space-based observatory (SBO) of mass $M$ has a circular orbital radius $R$ 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

$\frac{R}{4}$

B

$R$

C

$2R$

D

$4R$

Reveal Answer

A

$\frac{R}{4}$

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

$R$

Correct Answer

The orbital speed $v = \sqrt{\frac{G M_E}{R}}$ 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 $R$ .

C

$2R$

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

$4R$

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.

Q3

2021

VCAA

1 mark

Q3

1 mark

A 45 g golf ball, initially at rest, is hit by a golf club. The contact time between the club and the ball is 0.50 ms. The magnitude of the final velocity of the ball is 41 m s $^{-1}$ .

Which one of the following is closest to the average force experienced by the golf ball?

A

0.18 kN

B

0.37 kN

C

1.8 kN

D

3.7 kN

Reveal Answer

A

0.18 kN

This is incorrect. This value does not correspond to the correct calculation of average force using the impulse-momentum theorem.

B

0.37 kN

This is incorrect. This result likely comes from a unit conversion error, such as misinterpreting the mass or time units by a factor of 10.

C

1.8 kN

This is incorrect. This value represents the impulse or change in momentum ( $1.845 \text{ N s}$ ), not the average force.

D

3.7 kN

Correct Answer

This is correct. The average force is the change in momentum divided by time: $F = \frac{0.045 \text{ kg} \times 41 \text{ m s}^{-1}}{0.50 \times 10^{-3} \text{ s}} = 3690 \text{ N}$ , which is approximately $3.7 \text{ kN}$ .

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.

Q2

2025

VCAA

2 marks

Q2

2 marks

Airbags are designed to protect you in a crash by increasing the time interval, $\Delta t$ , over which your momentum changes when your car suddenly comes to a stop.

Explain, using physics principles, why this should reduce the severity of injuries for people in the car.

Reveal Answer

The severity of injuries will be directly proportional to the force applied.

Since $\Delta p = F \Delta t$ , if the time for the change in momentum is increased, the force can be decreased.

Marking Criteria

Descriptor	Marks
Identifies that the severity of injuries is directly proportional to the force applied.	1
Explains that for a given change in momentum, an increase in the time interval ( $\Delta t$ ) results in a decreased force ( $F$ ), referencing the relationship $\Delta p = F \Delta t$ or $F = \frac{\Delta p}{\Delta t}$ .	1

Q9

2022

QCAA

Paper 2

9 marks

Q9

A person spins an object 4.3 m above the ground in a horizontal circular path of radius 0.8 m. They release the object horizontally, allowing it to travel to the ground.

Q9a

4 marks

Calculate the centripetal acceleration of the object before it is released, given it takes 5 s for the object to complete 12 revolutions. Show your working. ( $ms^{-2}$ to two significant figures)

Reveal Answer

$v = \frac{2\pi r}{T}$
$= \frac{2\pi \times 0.8}{5 \div 12}$
$= 12.1\ m\ s^{-1}$

$a_c = \frac{v^2}{r}$
$= \frac{12.1^2}{0.8}$
$= 180\ m\ s^{-2}$

Centripetal acceleration = 180 m s $^{-2}$ (to two significant figures)

Marking Criteria

Descriptor	Marks
Recognises the scenario relates to velocity in circular motion	1
Recognises the scenario relates to centripetal acceleration	1
Provides appropriate mathematical reasoning	1
Calculates the centripetal acceleration of the object	1

Q9b

5 marks

Calculate the total horizontal displacement for the object after it is released. Show your working. (m to two significant figures)

Reveal Answer

$s_y = u_y t + \frac{1}{2} a t^2$
$4.3 = 0 + \frac{1}{2} \times 9.8 t^2$
$t = \sqrt{\frac{4.3}{4.9}}$
$= 0.94\ s$

$s_x = u_x t + \frac{1}{2} a t^2$
$= 12.1 \times 0.94 + \frac{1}{2} \times 0 \times 0.94^2$
$= 11.4\ m$

Horizontal displacement = 11 m (to two significant figures)

Marking Criteria

Descriptor	Marks
Recognises the scenario relates to vertical component of projectile motion	1
Provides appropriate mathematical reasoning	1
Determines the time of flight	1
Recognises the scenario relates to the horizontal component of projectile motion	1
Calculates the total horizontal displacement	1

Q7

2022

VCAA

1 mark

Q7

1 mark

A car travelling at $60 \text{ km h}^{-1}$ brakes to a complete stop over a distance of 18 m under a constant braking force.
Which one of the following is closest to the braking distance required for the same car to come to a complete stop when travelling at $40 \text{ km h}^{-1}$ and braking with the same constant braking force?

A

8 m

B

9 m

C

12 m

D

15 m

Reveal Answer

A

8 m

Correct Answer

Braking distance is proportional to the square of the initial velocity ( $s \propto u^2$ ). Since the speed is reduced by a factor of $2/3$ (from 60 to 40), the braking distance is multiplied by $(2/3)^2 = 4/9$ , giving $18 \times 4/9 = 8 \text{ m}$ .

B

9 m

This is half the original distance, which would only be correct if the speed was reduced by a factor of $1/\sqrt{2}$ , not $2/3$ .

C

12 m

This incorrectly assumes braking distance is directly proportional to velocity ( $18 \times 2/3 = 12 \text{ m}$ ). Kinetic energy and braking distance are proportional to velocity squared.

D

15 m

This value does not follow the $s \propto u^2$ relationship required by the kinematic equations for constant deceleration.

Q25

2020

QCAA

Paper 1

3 marks

Q25

3 marks

Mars has an average orbital radius of approximately 1.5 times the average orbital radius of Earth.
Calculate the time it takes Mars to orbit the Sun. (days to the nearest whole number)

Reveal Answer

$\frac{T_{\mathrm{Mars}}^2}{r_{\mathrm{Mars}}^3}=\frac{4\pi^2}{GM}$

$T=\sqrt{\frac{4\pi^2}{GM}\,r^3}$

$T_{\mathrm{Mars}}=\sqrt{\frac{4\pi^2}{GM}\,(1.5 \times r_{\mathrm{Earth}})^3}$

Therefore $T_{\mathrm{Mars}}=\sqrt{1.5^3}T_{\mathrm{Earth}}$

$T_{\mathrm{Mars}} = 1.8371 \times 365$ days

$T_{\mathrm{Mars}} = 670.5$ days

Time = 671 days

Marking Criteria

Descriptor	Marks
Indicates an understanding of the physical scenario in relation to Kepler’s law (or other relevant physical concept/s).	1
Provides pertinent mathematical operation/s correctly performed.	1
Determines the time correctly (accept: 1.83 to 1.84 years inclusive; OR 670 to 671 days inclusive; OR 16 080 to 16 107 hours inclusive; OR 57 888 000 to 57 974 400 seconds inclusive). Allow follow-through (FT) error for the time.	1

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