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QCAA Physics — Gravity and motion Questions & Answers

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

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 $2\pi$ .

Q7

2022

QCAA

Paper 2

3 marks

Q7

3 marks

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

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

Reveal Answer

$F = \frac{GMm}{r^2}$
$3.3 \times 10^3 = \frac{6.67 \times 10^{-11} \times 2.7 \times 10^{17} \times 6.1 \times 10^{15}}{r^2}$
$r = \sqrt{\frac{6.67 \times 10^{-11} \times 2.7 \times 10^{17} \times 6.1 \times 10^{15}}{3.3 \times 10^3}}$
$= 5.8 \times 10^9\ m$

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

Marking Criteria

Descriptor	Marks
Recognises the scenario relates to Newton’s Law of Universal Gravitation	1
Provides appropriate mathematical reasoning	1
Calculates the distance between the asteroids	1

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.

Q4

2023

QCAA

Paper 2

3 marks

Q4

3 marks

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

Object	Mass (kg)	Acceleration due to gravity (m s $^{-2}$ )
A	79	1.6
B	32	3.7

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

Reveal Answer

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

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

Object A experiences the greatest force.

Marking Criteria

Descriptor	Marks
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

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 ( $T^2 \propto r^3$ ) is physically derived by equating the centripetal force in uniform circular motion to the gravitational force defined by Newton's Law of Universal Gravitation.

Q12

2021

QCAA

Paper 1

1 mark

Q12

1 mark

Calculate the maximum height reached by a projectile with an initial velocity of 15 m s $^{-1}$ at an angle of 30° up from the horizontal.

A

2.87 m

B

3.83 m

C

8.61 m

D

11.5 m

Reveal Answer

A

2.87 m

Correct Answer

This is the correct maximum height derived from the formula $H = \frac{v^2 \sin^2(\theta)}{2g}$ . Substituting the values yields $\frac{(15)^2 (0.5)^2}{2(9.8)} \approx 2.87 \text{ m}$ .

B

3.83 m

This option is incorrect; it may result from a calculation error or using an incorrect fraction for the vertical component.

C

8.61 m

This answer incorrectly uses the horizontal component (cosine) instead of the vertical component (sine), calculating $\frac{v^2 \cos^2(\theta)}{2g}$ .

D

11.5 m

This option ignores the launch angle and calculates the height as if the projectile were fired straight up ( $90^\circ$ ) using $\frac{v^2}{2g}$ .

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 $mv^2/r = R$ .

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

Q24

2025

NESA

3 marks

Q24

3 marks

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

Show that the mass of $A$ is three times the mass of $B$ .

Reveal Answer

\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 $r_A = 3r_B$ :

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

$\therefore m_A = 3m_B$

Marking Criteria

Descriptor	Marks
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

Q10

2020

SCSA

6 marks

Q10

6 marks

A golfer hits a ball at 37.0 $\text{m s}^{-1}$ at 31.0° to the horizontal on a flat fairway. It travels 123 m. She wants to hit a target 135 m away, so she increases the angle at which she hits the ball, without changing the launch speed. Calculate the smallest increase of angle that allows her to reach the target. (Hint: $2\sin\theta\cos\theta = \sin 2\theta$ )

Reveal Answer

$t = 135/37 \cos\Theta$
$0 = 37 \sin\Theta - 4.9 (135/37 \cos\Theta)$
$37^2\sin\Theta\cos\Theta = 4.9 \times 135$
$\sin2\Theta = 2 \times 4.9 \times 135/37^2$
$2\Theta = 75.1^\circ$
$\Theta = 37.5^\circ$
$37.5 - 31 = 6.5^\circ$

Marking Criteria

Expresses t as range over horizontal velocity

Marking Bands

Descriptor	Marks
expresses t as range over horizontal velocity (t = 135/37 cosΘ)	1
None of the above	0

Substitutes time into equation for vertical displacement

Marking Bands

Descriptor	Marks
substitutes time into equation for vertical displacement (s = 0) and simplifies (0 = 37 sinΘ - 4.9 (135/37 cosΘ), 37²sinΘcosΘ = 4.9 × 135)	2
substitutes time into equation for vertical displacement (s = 0) (0 = 37 sinΘ - 4.9 (135/37 cosΘ))	1
None of the above	0

Solves for angle using expression given

Marking Bands

Descriptor	Marks
solves for angle using expression given (sin2Θ = 2 × 4.9 × 135/37², 2Θ = 75.1°, Θ = 37.5°)	2
partially solves for angle using expression given (e.g., sin2Θ = 2 × 4.9 × 135/37²)	1
None of the above	0

Subtracts initial angle to find change of angle

Marking Bands

Descriptor	Marks
subtracts initial angle to find change of angle (37.5 - 31 = 6.5°)	1
None of the above	0

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.

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 $\sqrt{M/R}$ .

Q2

2021

QCAA

Paper 1

1 mark

Q2

1 mark

Calculate the initial horizontal velocity of a projectile with an initial velocity of 38 m s $^{-1}$ at an angle of 42° up from the horizontal.

A

25 m s $^{-1}$

B

28 m s $^{-1}$

C

34 m s $^{-1}$

D

40 m s $^{-1}$

Reveal Answer

A

25 m s $^{-1}$

This is the initial vertical velocity, calculated using $v_y = v \sin(\theta) = 38 \sin(42^\circ) \approx 25 \text{ m s}^{-1}$ .

B

28 m s $^{-1}$

Correct Answer

The horizontal component of velocity is found using the cosine of the angle: $v_x = v \cos(\theta) = 38 \cos(42^\circ) \approx 28.2 \text{ m s}^{-1}$ .

C

34 m s $^{-1}$

This value is incorrect; it does not result from applying the trigonometric component formulas to the given velocity and angle.

D

40 m s $^{-1}$

This is impossible because a component of the velocity vector ( $v_x$ ) cannot be greater than the total magnitude of the velocity ( $38 \text{ m s}^{-1}$ ).

Q22

2023

QCAA

Paper 1

3 marks

Q22

3 marks

Particles move at a rate of $1.3\times10^6$ 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 = ______ $ms^{-1}$ (two significant figures)

Reveal Answer

$C = 2\pi r = 2 \times \pi \times 35 \approx 219.91 \text{ m}$

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

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

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

Marking Criteria

Descriptor	Marks
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

Q27

2023

QCAA

Paper 1

5 marks

Q27

5 marks

A satellite orbits a planet of mass $6.42\times 10^{23}$ 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 = _____ $ms^{-1}$ (two significant figures)

Reveal Answer

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

$\frac{T^2}{(3390 \times 10^3 + 5000 \times 10^3)^3} = \frac{4\pi^2}{6.67 \times 10^{-11} \times 6.42 \times 10^{23}}$

$T = \sqrt{\frac{4\pi^2 \times 5.91 \times 10^{20}}{6.67 \times 10^{-11} \times 6.42 \times 10^{23}}}$
$= 2.33 \times 10^4 \text{ s}$

$v = \frac{2\pi r}{T}$
$= \frac{2\pi \times 8.39 \times 10^6}{2.33 \times 10^4}$
$= 2.3 \times 10^3 \text{ m s}^{-1}$

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

Marking Criteria

Descriptor	Marks
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

QCAA Physics Gravity and motion

Expresses t as range over horizontal velocity

Substitutes time into equation for vertical displacement

Solves for angle using expression given

Subtracts initial angle to find change of angle

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QCAA Physics Gravity and motion

Sample Answer

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Expresses t as range over horizontal velocity

Substitutes time into equation for vertical displacement

Solves for angle using expression given

Subtracts initial angle to find change of angle

Sample Answer

Sample Answer

Frequently Asked Questions

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