AQA AS Paper 1 2023 June — Question 17 4 marks

Exam BoardAQA
ModuleAS Paper 1 (AS Paper 1)
Year2023
SessionJune
Marks4
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicVectors Introduction & 2D
TypeResultant of two vector forces (direction/magnitude conditions)
DifficultyModerate -0.8 This is a straightforward AS-level mechanics question requiring basic vector operations: magnitude calculation using Pythagoras (part a), finding a displacement vector by subtraction (part b(i)), and using parallel vectors to set up a simple proportion (part b(ii)). All techniques are routine with no problem-solving insight needed, making it easier than average but not trivial due to the multi-part structure.
Spec1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication1.10e Position vectors: and displacement
A particle, \(P\), is initially at rest on a smooth horizontal surface. A resultant force of \(\begin{bmatrix} 12 \\ 9 \end{bmatrix}\) N is then applied to \(P\), so that it moves in a straight line.
  1. Find the magnitude of the resultant force. [1 mark]
  2. Two fixed points \(A\) and \(B\) have position vectors $$\overrightarrow{OA} = \begin{bmatrix} 3 \\ 7 \end{bmatrix} \text{ metres} \quad \text{and} \quad \overrightarrow{OB} = \begin{bmatrix} k \\ k-1 \end{bmatrix} \text{ metres}$$ with respect to a fixed origin, \(O\) \(P\) moves in a straight line parallel to \(\overrightarrow{AB}\)
    1. Find \(\overrightarrow{AB}\) in terms of \(k\) [1 mark]
    2. Find the value of \(k\) [2 marks]
Question 17:
AnswerMarks
17(a)Finds correct magnitude of
force.
AnswerMarks Guidance
Condone omission of units1.1b B1
Subtotal1
QMarking instructions AO
AnswerMarks
17(b)(i)Forms correct expression for
AB
AnswerMarks Guidance
Condone omission of units1.1b B1
A B = metres
k − 8
AnswerMarks Guidance
Subtotal1
QMarking instructions AO
AnswerMarks
17(b)(ii)Deduces A B is a scalar multiple
 1 2 
of
AnswerMarks Guidance
92.2a M1
direction of force then A B is a
 1 2 
scalar multiple of
9
k−3 12
=
   
k−8 9
9(k – 3) = 12(k – 8)
9k – 27 = 12k – 96
k = 23
Deduces k = 23
AnswerMarks Guidance
FT their answer to part (b)(i)2.2a A1F
Subtotal2
Question 17 Total4
QMarking instructions AO 
Question 17:
--- 17(a) ---
17(a) | Finds correct magnitude of
force.
Condone omission of units | 1.1b | B1 | 1 2 2 + 9 2 = 1 5 N
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 17(b)(i) ---
17(b)(i) | Forms correct expression for
AB
Condone omission of units | 1.1b | B1 |  k − 3 
A B = metres
k − 8
Subtotal | 1
Q | Marking instructions | AO | Marks | Typical solution
--- 17(b)(ii) ---
17(b)(ii) | Deduces A B is a scalar multiple
 1 2 
of
9 | 2.2a | M1 | Since direction of movement is in
direction of force then A B is a
 1 2 
scalar multiple of
9
k−3 12
=
   
k−8 9
9(k – 3) = 12(k – 8)
9k – 27 = 12k – 96
k = 23
Deduces k = 23
FT their answer to part (b)(i) | 2.2a | A1F
Subtotal | 2
Question 17 Total | 4
Q | Marking instructions | AO | Marks | Typical solution
A particle, $P$, is initially at rest on a smooth horizontal surface.

A resultant force of $\begin{bmatrix} 12 \\ 9 \end{bmatrix}$ N is then applied to $P$, so that it moves in a straight line.

\begin{enumerate}[label=(\alph*)]
\item Find the magnitude of the resultant force.
[1 mark]

\item Two fixed points $A$ and $B$ have position vectors
$$\overrightarrow{OA} = \begin{bmatrix} 3 \\ 7 \end{bmatrix} \text{ metres} \quad \text{and} \quad \overrightarrow{OB} = \begin{bmatrix} k \\ k-1 \end{bmatrix} \text{ metres}$$
with respect to a fixed origin, $O$

$P$ moves in a straight line parallel to $\overrightarrow{AB}$

\begin{enumerate}[label=(\roman*)]
\item Find $\overrightarrow{AB}$ in terms of $k$
[1 mark]

\item Find the value of $k$
[2 marks]
\end{enumerate}
\end{enumerate}

\hfill \mbox{\textit{AQA AS Paper 1 2023 Q17 [4]}}
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