Question 16 - A-Level Maths

AQA Paper 2 2023 June — Question 16 4 marks

Exam Board	AQA
Module	Paper 2 (Paper 2)
Year	2023
Session	June
Marks	4
Paper	Download PDF ↗
Mark scheme	Download PDF ↗
Topic	Vectors Introduction & 2D
Type	Newton's second law with vector forces (find acceleration or force)
Difficulty	Moderate -0.8 This is a straightforward application of Newton's second law (F=ma) with vector addition. Students add the two force vectors, equate to mass times acceleration, then solve a simple linear equation for k. It requires only routine vector arithmetic and basic algebraic manipulation, making it easier than average for A-level.
Spec	1.10d Vector operations: addition and scalar multiplication 3.03d Newton's second law: 2D vectors

16 A particle moves under the action of two forces, $\mathbf { F } _ { 1 }$ and $\mathbf { F } _ { 2 }$ It is given that $$\begin{aligned} & \mathbf { F } _ { 1 } = ( 1.6 \mathbf { i } - 5 \mathbf { j } ) \mathrm { N } \\ & \mathbf { F } _ { 2 } = ( k \mathbf { i } + 5 k \mathbf { j } ) \mathrm { N } \end{aligned}$$ where $k$ is a constant.
The acceleration of the particle is $( 3.2 \mathbf { i } + 12 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 2 }$ Find $k$ \includegraphics[max width=\textwidth, alt={}, center]{de8a7d38-a665-4feb-854e-ac83f413d133-25_2488_1716_219_153}

Show mark scheme Show mark scheme source

Question 16:

Answer	Marks	Guidance
$\mathbf{F}_1 + \mathbf{F}_2 = \begin{bmatrix}1.6+k\\5k-5\end{bmatrix}$	B1	Adds the two forces together; ACF
$\begin{bmatrix}1.6+k\\5k-5\end{bmatrix} = m\begin{bmatrix}3.2\\12\end{bmatrix}$	M1	Uses $\mathbf{F}=m\mathbf{a}$ and substitutes $\mathbf{F}_1 \pm \mathbf{F}_2$ and $\mathbf{a}=\begin{bmatrix}3.2\\12\end{bmatrix}$
$1.6+k = 3.2m$ and $5k-5=12m$	A1	Obtains two correct equations (only award if vectors removed); or obtains correct linear equation in $k$, e.g. $\frac{5k-5}{12}=\frac{1.6+k}{3.2}$
$k = 8.8$	A1	ACF

## Question 16:

$\mathbf{F}_1 + \mathbf{F}_2 = \begin{bmatrix}1.6+k\\5k-5\end{bmatrix}$ | B1 | Adds the two forces together; ACF

$\begin{bmatrix}1.6+k\\5k-5\end{bmatrix} = m\begin{bmatrix}3.2\\12\end{bmatrix}$ | M1 | Uses $\mathbf{F}=m\mathbf{a}$ and substitutes $\mathbf{F}_1 \pm \mathbf{F}_2$ and $\mathbf{a}=\begin{bmatrix}3.2\\12\end{bmatrix}$

$1.6+k = 3.2m$ and $5k-5=12m$ | A1 | Obtains two correct equations (only award if vectors removed); or obtains correct linear equation in $k$, e.g. $\frac{5k-5}{12}=\frac{1.6+k}{3.2}$

$k = 8.8$ | A1 | ACF

---

Show LaTeX source

16 A particle moves under the action of two forces, $\mathbf { F } _ { 1 }$ and $\mathbf { F } _ { 2 }$

It is given that

$$\begin{aligned}
& \mathbf { F } _ { 1 } = ( 1.6 \mathbf { i } - 5 \mathbf { j } ) \mathrm { N } \\
& \mathbf { F } _ { 2 } = ( k \mathbf { i } + 5 k \mathbf { j } ) \mathrm { N }
\end{aligned}$$

where $k$ is a constant.\\
The acceleration of the particle is $( 3.2 \mathbf { i } + 12 \mathbf { j } ) \mathrm { m } \mathrm { s } ^ { - 2 }$\\
Find $k$\\

\includegraphics[max width=\textwidth, alt={}, center]{de8a7d38-a665-4feb-854e-ac83f413d133-25_2488_1716_219_153}

\hfill \mbox{\textit{AQA Paper 2 2023 Q16 [4]}}

This paper (20 questions)

View full paper

Q1 1 Q2 1 Q3 1 Q4 7 Q5 7 Q6 6 Q7 5 Q8 10 Q9 6 Q10 6 Q11 1 Q12 1 Q13 5 Q14 4 Q15 4 Q16 4 Q17 6 Q18 6 Q19 12 Q20 7

Answer	Marks	Guidance
\(\mathbf{F}_1 + \mathbf{F}_2 = \begin{bmatrix}1.6+k\\5k-5\end{bmatrix}\)	B1	Adds the two forces together; ACF
\(\begin{bmatrix}1.6+k\\5k-5\end{bmatrix} = m\begin{bmatrix}3.2\\12\end{bmatrix}\)	M1	Uses \(\mathbf{F}=m\mathbf{a}\) and substitutes \(\mathbf{F}_1 \pm \mathbf{F}_2\) and \(\mathbf{a}=\begin{bmatrix}3.2\\12\end{bmatrix}\)
\(1.6+k = 3.2m\) and \(5k-5=12m\)	A1	Obtains two correct equations (only award if vectors removed); or obtains correct linear equation in \(k\), e.g. \(\frac{5k-5}{12}=\frac{1.6+k}{3.2}\)
\(k = 8.8\)	A1	ACF