Specialist Mathematics · Past QCAA questions All simple familiar questions
2025 · Paper 2 · Simple familiar

Question 12

[7 marks] Technology-active
Topic guidePremiumUnit 4 Topic 4.2 · Forces and momentum
Game plansPremium8.1 Drawing the force diagram8.2 Choosing the acceleration form
The question

A 10 kg object is travelling at ground level with a constant velocity.

At an instant, two forces of 60 N and 42 N act simultaneously on the object parallel to the ground in the directions shown.

NorthSouthEastWestNot to scale 18°32°42 N60 N

Let unit vectors in the east and north directions be $\hat{\mathbf{i}}$ and $\hat{\mathbf{j}}$ respectively.

a)

Determine the resultant force acting on the object expressed in Cartesian form. Simplify your answer.

[2 marks] b)

Determine the acceleration of the object after the forces act. Leave your answer in Cartesian form.

[1 mark]

The object is initially travelling at $5\ \text{m s}^{-1}$ in a northerly direction when the forces act.

c)

Determine an expression for the object’s velocity in terms of time, $t$, after the forces act. Leave your answer in Cartesian form.

[2 marks] d)

Calculate the speed of the object after the forces have been acting for two seconds.

[2 marks]
Watch the situation first

The resultant pulls south-west, so the object curves away

Watch the two forces break into east and north parts and combine into one orange resultant. Then watch where that sends the object.

Two forces act on a 10 kg object moving north Resolve each force into east and north parts Add the parts to get the resultant, then divide by 10 The object slows and curves to the west After 2 seconds its speed is about 2.51 m s−1 NorthSouthEastWest 5 m s−1 18° 42 N 32° 60 N resultant East 42 cos 18° − 60 cos 32° ≈ −10.94 North 42 sin 18° − 60 sin 32° ≈ −18.82 F ≈ −10.94 î − 18.82 ĵ N a = F ÷ 10 ≈ −1.094 î − 1.882 ĵ North start 2.51 m s−1 t = 2 s

The 42 N force has positive east and north parts. The 60 N force points west and south, so both of its parts are negative.

Now for the mathematics

Move through time and watch the velocity change

Move the slider to change $t$. The orange arrow is the velocity, which always runs along the path. The grey arrow is the acceleration, which never changes.

−8−6−4−2246 East (m)North (m)

At $t=0$ the velocity is $5\hat{\mathbf{j}}$, straight north. This is the initial condition you use to find the constant in part c).

This is the time in part d). The velocity and acceleration are almost at right angles here, which is why the object is close to its slowest.

The acceleration never changes, but the velocity does. The object slows down until about $t=2$, then speeds up again as it heads south-west.

QCAA marking guide · 7 marks

Work through the solution one mark at a time

Try each step yourself before you reveal it. Keep the unrounded values in your calculator until the last line.

Prefer to watch? Watch me work through it on YouTube (opens in a new tab)

Step 1 · part a)1 mark

Resolve each force using the angle it makes with its nearest axis. The 60 N force points west and south, so both of its parts are negative.

$$\mathbf{F}_{\text{net}}=\left(42\cos18^\circ\,\hat{\mathbf{i}}+42\sin18^\circ\,\hat{\mathbf{j}}\right)+\left(-60\cos32^\circ\,\hat{\mathbf{i}}-60\sin32^\circ\,\hat{\mathbf{j}}\right)$$

Using angles from the positive $x$-axis also works: $60\cos212^\circ\,\hat{\mathbf{i}}+60\sin212^\circ\,\hat{\mathbf{j}}$ gives the same parts.

Marker: correctly resolves the given vectors

Step 2 · part a)1 mark

Collect the $\hat{\mathbf{i}}$ parts and the $\hat{\mathbf{j}}$ parts.

$$\begin{aligned}\hat{\mathbf{i}}:&\quad 39.944-50.883=-10.939\\ \hat{\mathbf{j}}:&\quad 12.979-31.795=-18.816\end{aligned}$$

$\mathbf{F}_{\text{net}}\approx-10.94\,\hat{\mathbf{i}}-18.82\,\hat{\mathbf{j}}$ N

Your answer has to be a vector. A magnitude and direction alone would not be in Cartesian form.

Marker: determines a simplified expression for the resultant force in Cartesian form

Step 3 · part b)1 mark

Use $\mathbf{F}_{\text{net}}=m\mathbf{a}$ with $m=10$. Divide each part by 10.

$$\mathbf{a}=\frac{1}{10}\left(-10.94\,\hat{\mathbf{i}}-18.82\,\hat{\mathbf{j}}\right)$$

$\mathbf{a}\approx-1.094\,\hat{\mathbf{i}}-1.882\,\hat{\mathbf{j}}\ \text{m s}^{-2}$

Marker: determines acceleration in Cartesian form

Step 4 · part c)1 mark

Velocity is the integral of acceleration. Each part is a constant, so each integrates to a multiple of $t$.

$$\mathbf{v}=\int\mathbf{a}\,dt=-1.094t\,\hat{\mathbf{i}}-1.882t\,\hat{\mathbf{j}}+\mathbf{c}$$

The constant is a vector, because each part has its own constant. This mark can be implied by your next line.

Marker: determines expression for velocity

Step 5 · part c)1 mark

At $t=0$, the object is moving north at $5\ \text{m s}^{-1}$, so $\mathbf{v}=5\hat{\mathbf{j}}$. That gives $\mathbf{c}=5\hat{\mathbf{j}}$.

$\mathbf{v}=-1.094t\,\hat{\mathbf{i}}+(5-1.882t)\,\hat{\mathbf{j}}$

Marker: determines velocity based on initial condition in Cartesian form

Step 6 · part d)1 mark

Substitute $t=2$.

$$\mathbf{v}(2)=-2.188\,\hat{\mathbf{i}}+1.237\,\hat{\mathbf{j}}\approx-2.19\,\hat{\mathbf{i}}+1.24\,\hat{\mathbf{j}}$$

The QCAA accept rounded values such as $-2.2\hat{\mathbf{i}}+1.2\hat{\mathbf{j}}$.

Marker: determines velocity when $t=2$

Step 7 · part d)1 mark

Speed is the magnitude of the velocity.

$$|\mathbf{v}(2)|=\sqrt{(-2.188)^2+(1.237)^2}$$

$|\mathbf{v}(2)|\approx2.51\ \text{m s}^{-1}$

A rounded value of 2.5 is also accepted. The question asks for speed, so give a single number. A vector here would not answer the question.

Marker: determines speed when $t=2$

Putting it all together

Each phrase of the question gave you something

“A 10 kg object”
This is $m$ in $\mathbf{F}=m\mathbf{a}$.
“with a constant velocity”
Before the two forces act, the net force is zero. After they act, they are the whole resultant.
“unit vectors in the east and north directions”
East and north are positive. West and south parts get a minus sign.
“parallel to the ground”
Everything happens in two dimensions, so you only need $\hat{\mathbf{i}}$ and $\hat{\mathbf{j}}$.
“initially travelling at $5\ \text{m s}^{-1}$ in a northerly direction”
This is $\mathbf{v}(0)=5\hat{\mathbf{j}}$, which finds the constant.
“Calculate the speed”
Speed is the magnitude of the velocity, a single positive number.
What makes this simple familiar

Each part is one routine step from the forces topic. You resolve and add, divide by the mass, integrate with an initial condition, then take a magnitude. Each part uses your answer from the one before.

If I had to guess where you might lose marks, it would be the signs on the 60 N force in part a). It points west and south, so both of its parts are negative. In part c), forgetting the $5\hat{\mathbf{j}}$ costs you the second mark. In part d), give the speed as a number. Stopping at the velocity vector loses a mark.

Only mark this done when you could do it without help. Reading the solution does not count.

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Question wording and marking-guide steps are from the 2025 QCAA Specialist Mathematics external assessment, © State of Queensland (QCAA) 2025, licensed under CC BY 4.0, and have been adapted. Tangent Tuition is not affiliated with the QCAA.

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