QCAA Physics Alternative Sequence Linear motion and force

15 sample questions with marking guides and sample answers

Q20
2021
QCAA
Paper 1
1 mark
Q20
1 mark

To determine a value for acceleration due to gravity, a student dropped an object from a height equal to their eye-level and counted the time from release to impact. The student then conducted the experiment two more times.

To reduce the percentage error of the experimental value for acceleration due to gravity, the student should

A

use an object with a lower mass.

B

conduct more trials of the experiment.

C

increase the height from which the object is released.

D

use a stopwatch to measure the time from release to impact.

Reveal Answer
A

use an object with a lower mass.

Acceleration due to gravity is independent of an object's mass, so using a lighter or heavier object will not affect the experimental value or its percentage error.

B

conduct more trials of the experiment.

While conducting more trials reduces random error and improves the reliability of the average result, it does not decrease the percentage error inherent in the individual measurements themselves.

C

increase the height from which the object is released.

Correct Answer

Increasing the drop height increases the total time of the fall. Since the absolute uncertainty in timing (like human reaction time) remains relatively constant, a larger measured time results in a significantly smaller percentage error.

D

use a stopwatch to measure the time from release to impact.

Although a stopwatch is more precise than counting, human reaction time (typically around 0.2 s0.2\text{ s}) still introduces a massive percentage error for a short drop from eye-level (which takes only about 0.5 s0.5\text{ s}).

Q1
2021
QCAA
Paper 1
0 marks
Q1b
0 marks

Leptons do not experience the

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.

Q20
2025
VCAA
1 mark
Q20
1 mark

Harriet and Tom were investigating how the speed, vv, of a falling object varied with the distance, ss, 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

vv versus ss

B

vv versus s\sqrt{s}

C

v2v^2 versus s\sqrt{s}

D

v\sqrt{v} versus ss

Reveal Answer
A

vv versus ss

The kinematic equation for an object falling from rest is v2=2gsv^2 = 2gs, meaning vv is proportional to s\sqrt{s}. A graph of vv versus ss would result in a curve, not a straight line.

B

vv versus s\sqrt{s}

Correct Answer

Using the kinematic equation v2=u2+2asv^2 = u^2 + 2as with an initial velocity of u=0u=0, we get v=2gsv = \sqrt{2g}\sqrt{s}. This shows that vv is directly proportional to s\sqrt{s}, which produces a straight line through the origin.

C

v2v^2 versus s\sqrt{s}

Based on the equation v2=2gsv^2 = 2gs, v2v^2 is directly proportional to ss, not s\sqrt{s}. Plotting v2v^2 versus s\sqrt{s} would result in a quadratic curve.

D

v\sqrt{v} versus ss

Since vv is proportional to s1/2s^{1/2}, v\sqrt{v} would be proportional to s1/4s^{1/4}. Plotting v\sqrt{v} versus ss would not produce a straight line.

Q4
2021
QCAA
Paper 1
1 mark
Q4
1 mark

What is the final velocity of a 5 kg5\ \text{kg} object dropped from a height of 16 m16\ \text{m}?

A

9 m s19\ \text{m s}^{-1}

B

11 m s111\ \text{m s}^{-1}

C

13 m s113\ \text{m s}^{-1}

D

18 m s118\ \text{m s}^{-1}

Reveal Answer
A

9 m s19\ \text{m s}^{-1}

Incorrect. This value is too low and does not follow the kinematic equation v=2ghv = \sqrt{2gh}.

B

11 m s111\ \text{m s}^{-1}

Incorrect. This velocity is incorrect; remember that the final velocity of a dropped object is independent of its mass and depends only on height and gravity.

C

13 m s113\ \text{m s}^{-1}

Incorrect. This is less than the actual final velocity calculated using v=2ghv = \sqrt{2gh}.

D

18 m s118\ \text{m s}^{-1}

Correct Answer

Correct. Using the kinematic equation v=2ghv = \sqrt{2gh} (where initial velocity is zero), the final velocity is 2×9.8×1617.7 m s1\sqrt{2 \times 9.8 \times 16} \approx 17.7\ \text{m s}^{-1}, which rounds to 18 m s118\ \text{m s}^{-1}. The mass of 5 kg5\ \text{kg} is extra information, as all objects fall at the same rate in a vacuum.

Q26
2021
QCAA
Paper 1
4 marks
Q26
4 marks

A 3.2 kg3.2\ \text{kg} object experiences an upwards force of 6.2 N6.2\ \text{N} as it falls through a viscous liquid in a cylindrical container.

Calculate the total work done on the object as it falls 13.6 cm13.6\ \text{cm} through the liquid (to 2 significant figures). Show your working.

Reveal Answer

Fg=mg=3.2×9.8=31.4 NF_g = mg = 3.2 \times 9.8 = 31.4\text{ N}

Fnet=FgFupwards=31.46.2=25.2 NF_{net} = F_g - F_{upwards} = 31.4 - 6.2 = 25.2\text{ N}

W=Fs=25.2×0.136W = Fs = 25.2 \times 0.136

Total work = 3.4 J (to 2 significant figures)

Marking Criteria
DescriptorMarks

calculates the force due to gravity

1

calculates the net force

1

identifies the relationship between work and net force

1

calculates the total work done

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
Q26
2023
QCAA
Paper 1
6 marks
Q26
6 marks

A four-wheeled cart starting at rest moves along a surface with a constant applied force of 55 N. It experiences a frictional force of 8 N on each wheel.

Determine the mass of the cart if it travels 60 m in 10 s. Show your working.

Reveal Answer

Total friction = 4×8=32 N4 \times 8 = 32 \text{ N}

Fnet=5532=+23 N\therefore F_{net} = 55 - 32 = +23 \text{ N}

s=ut+12at260=0+12×102×aa=1.2 m s2\begin{align*} s &= ut + \frac{1}{2}at^2\\ 60 &= 0 + \frac{1}{2} \times 10^2 \times a\\ a &= 1.2 \text{ m s}^{-2} \end{align*}

Since

a=Fnetm1.2=23mm=19 kg\begin{align*} a &= \frac{F_{net}}{m}\\ 1.2 &= \frac{23}{m}\\ m &= 19 \text{ kg} \end{align*}

Mass = 19 kg19 \text{ kg} (to the nearest whole number)

Marking Criteria
DescriptorMarks

recognises scenario relates to the addition of vectors

1

recognises scenario relates to linear motion with constant acceleration

1

recognises scenario relates to Newton's second law

1

calculates magnitude of net force

1

calculates the acceleration of the cart

1

determines the mass of the cart

1
Q8
2023
QCAA
Paper 1
1 mark
Q8
1 mark

Two objects of equal mass, X and Y, move forward in the same direction. Object X starts behind object Y and moves 40 m s140 \text{ m s}^{-1} faster. The two objects eventually collide, resulting in a combined object Z that continues to move forward at three times the original velocity of object Y.

Determine the final velocity of object Z.

A

20 m s120 \text{ m s}^{-1}

B

30 m s130 \text{ m s}^{-1}

C

40 m s140 \text{ m s}^{-1}

D

50 m s150 \text{ m s}^{-1}

Reveal Answer
A

20 m s120 \text{ m s}^{-1}

This is incorrect. This value might result from an algebraic error when solving the conservation of momentum equation, such as solving for 2v2v instead of 3v3v.

B

30 m s130 \text{ m s}^{-1}

Correct Answer

Correct. Using conservation of momentum, m(v+40)+mv=2m(3v)m(v + 40) + mv = 2m(3v), which simplifies to 2v+40=6v2v + 40 = 6v. Solving for vv gives 10 m s110 \text{ m s}^{-1}, making the final velocity 3v=30 m s13v = 30 \text{ m s}^{-1}.

C

40 m s140 \text{ m s}^{-1}

This is incorrect. This value represents the difference in initial velocities between object X and object Y, not the final velocity of the combined object Z.

D

50 m s150 \text{ m s}^{-1}

This is incorrect. This value represents the initial velocity of object X (10 m s1+40 m s110 \text{ m s}^{-1} + 40 \text{ m s}^{-1}), rather than the final velocity of object Z.

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}.

Q9
2023
QCAA
Paper 1
1 mark
Q9
1 mark

The displacement, ss (in metres), of an object undergoing deceleration was recorded over time, tt (in seconds). An equation describing the trend was developed.

s=(1.3±0.2)t2+(203±2)s = (-1.3 \pm 0.2)t^2 + (203 \pm 2)

Given the uncertainties in the equation, what is a valid displacement of the object after seven seconds?

A

267 m

B

215 m

C

194 m

D

146 m

Reveal Answer
A

267 m

Incorrect. This value likely comes from ignoring the negative sign on the first term (1.3(7)2+203=266.71.3(7)^2 + 203 = 266.7 m), which falls far outside the valid uncertainty range of 127.5127.5 m to 151.1151.1 m.

B

215 m

Incorrect. This value is outside the valid displacement range of 127.5127.5 m to 151.1151.1 m, possibly resulting from an arithmetic error or misapplying the uncertainty values.

C

194 m

Incorrect. This value likely comes from forgetting to square the time variable (1.3(7)+203=193.9-1.3(7) + 203 = 193.9 m), placing it outside the valid range of 127.5127.5 m to 151.1151.1 m.

D

146 m

Correct Answer

Correct. Substituting t=7t = 7 yields s=(1.3±0.2)(49)+(203±2)=139.3±11.8s = (-1.3 \pm 0.2)(49) + (203 \pm 2) = 139.3 \pm 11.8 m. The value 146146 m is the only option that falls within this valid range of 127.5127.5 m to 151.1151.1 m.

Q18
2023
QCAA
Paper 1
1 mark
Q18
1 mark

A 20 kg object is placed on an inclined plane with a slope of 35°. If the object experiences a frictional force of 40 N and no additional applied force, calculate its acceleration down the inclined plane.

A

3.6 m s23.6 \text{ m s}^{-2}

B

5.6 m s25.6 \text{ m s}^{-2}

C

6.0 m s26.0 \text{ m s}^{-2}

D

7.6 m s27.6 \text{ m s}^{-2}

Reveal Answer
A

3.6 m s23.6 \text{ m s}^{-2}

Correct Answer

Correct. The net force is the component of gravity down the slope minus friction (mgsin(35)40mg \sin(35^\circ) - 40). Dividing this net force (72.472.4 N) by the mass (2020 kg) gives an acceleration of 3.6 m s23.6 \text{ m s}^{-2}.

B

5.6 m s25.6 \text{ m s}^{-2}

Incorrect. This value represents the acceleration of the object if there were no friction acting against it (a=gsin(35)a = g \sin(35^\circ)).

C

6.0 m s26.0 \text{ m s}^{-2}

Incorrect. This result comes from incorrectly using cosine instead of sine to find the component of gravity parallel to the slope (mgcos(35)mg \cos(35^\circ)).

D

7.6 m s27.6 \text{ m s}^{-2}

Incorrect. This calculates the acceleration if friction were incorrectly added to the gravitational force, acting in the same direction down the slope rather than opposing the motion.

Q20
2020
QCAA
Paper 1
1 mark
Q20
1 mark

Calculate the frequency of light that would be required to eject a photoelectron at a velocity of 1.90×106 m s11.90 \times 10^6 \text{ m s}^{-1} from a metal plate with a work function of 4.73 eV.

A

1.14×10151.14 \times 10^{15} Hz

B

1.34×10151.34 \times 10^{15} Hz

C

2.48×10152.48 \times 10^{15} Hz

D

3.62×10153.62 \times 10^{15} Hz

Reveal Answer
A

1.14×10151.14 \times 10^{15} Hz

This is the threshold frequency, calculated using only the work function (Φ/h\Phi/h), which ignores the additional energy needed for the kinetic energy of the ejected electron.

B

1.34×10151.34 \times 10^{15} Hz

This frequency incorrectly corresponds to the difference between the kinetic energy and the work function, rather than their sum.

C

2.48×10152.48 \times 10^{15} Hz

This frequency corresponds only to the kinetic energy of the electron (K/hK/h), ignoring the energy required to overcome the metal's work function.

D

3.62×10153.62 \times 10^{15} Hz

Correct Answer

The total photon energy is the sum of the work function and the electron's kinetic energy (E=Φ+12mv2E = \Phi + \frac{1}{2}mv^2). Dividing this total energy by Planck's constant gives the correct frequency.

Q14
2021
QCAA
Paper 1
1 mark
Q14
1 mark

Which example describes one of Newton’s laws of motion?

A

The acceleration of an object must always be uniform.

B

In the absence of a net force, an object maintains a constant velocity.

C

To an observer at rest, the length of a moving object appears shorter in the direction it is moving.

D

The force of attraction between each pair of point particles is inversely proportional to the square of the distance between them.

Reveal Answer
A

The acceleration of an object must always be uniform.

Incorrect. Newton's second law (F=maF = ma) shows that acceleration depends on the net force, which can change over time, meaning acceleration does not have to be uniform.

B

In the absence of a net force, an object maintains a constant velocity.

Correct Answer

Correct. This accurately describes Newton's first law of motion, also known as the law of inertia, where an object's velocity remains constant unless acted upon by a net external force.

C

To an observer at rest, the length of a moving object appears shorter in the direction it is moving.

Incorrect. This describes length contraction, a principle from Einstein's theory of special relativity, not classical Newtonian mechanics.

D

The force of attraction between each pair of point particles is inversely proportional to the square of the distance between them.

Incorrect. This describes Newton's law of universal gravitation, which is a separate physical principle from his three laws of motion.

Q1
2022
SCSA
5 marks
Q1
5 marks

Light with a wavelength of 341 nm is shone onto a potassium metal plate in a photoelectric cell, causing a photocurrent to flow. The work function of potassium is 2.30 eV. Calculate the maximum speed of the electrons emitted by the plate.

Reveal Answer

Using photoelectric effect equation: E=hfWE = hf - W

Convertine eV to Joules: 2.30×1.60×1019=3.68×1019 J2.30 \times 1.60 \times 10^{-19} = 3.68 \times 10^{-19} \text{ J}

Calculating energy of incoming photon:
E=hcλ=3.00×108×6.63×1034/341×109E = \frac{hc}{\lambda} = 3.00 \times 10^8 \times 6.63 \times 10^{-34} / 341 \times 10^{-9}
=5.83×1019 J= 5.83 \times 10^{-19} \text{ J}

Calculating kinetic energy of electron:
KEe=(5.833.68)×1019=2.15×1019 JKE_e = (5.83 - 3.68) \times 10^{-19} = 2.15 \times 10^{-19} \text{ J}

Velocity of electron:

v=2×2.15×10199.11×1031=6.87×105 m s1v = \sqrt{\frac{2 \times 2.15 \times 10^{-19}}{9.11 \times 10^{-31}}} = 6.87 \times 10^5 \text{ m s}^{-1}
Marking Criteria
DescriptorMarks

Uses photoelectric effect equation

1

Converts eV to Joules

1

Calculates the energy of incoming photon.

1

Calculates kinetic energy of electron

1

Calculates the velocity of the electron

1

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