AQA Paper 2 Specimen — Question 13 5 marks

Exam BoardAQA
ModulePaper 2 (Paper 2)
SessionSpecimen
Marks5
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Mark schemeDownload PDF ↗
TopicVectors Introduction & 2D
TypeResultant of three coplanar forces
DifficultyModerate -0.8 This is a straightforward mechanics question requiring only vector addition, Pythagoras' theorem, and basic trigonometry. All steps are routine: sum components, find magnitude, find angle using tan^(-1), then negate for equilibrium. No problem-solving insight needed, just direct application of standard formulas.
Spec1.10a Vectors in 2D: i,j notation and column vectors1.10c Magnitude and direction: of vectors1.10d Vector operations: addition and scalar multiplication3.03a Force: vector nature and diagrams3.03m Equilibrium: sum of resolved forces = 0
The three forces \(\mathbf{F_1}\), \(\mathbf{F_2}\) and \(\mathbf{F_3}\) are acting on a particle. \(\mathbf{F_1} = (25\mathbf{i} + 12\mathbf{j})\) N \(\mathbf{F_2} = (-7\mathbf{i} + 5\mathbf{j})\) N \(\mathbf{F_3} = (15\mathbf{i} - 28\mathbf{j})\) N The unit vectors \(\mathbf{i}\) and \(\mathbf{j}\) are horizontal and vertical respectively. The resultant of these three forces is \(\mathbf{F}\) newtons.
  • The fourth force, \(\mathbf{F_4}\), is applied to the particle so that the four forces are in equilibrium. Find \(\mathbf{F_4}\), giving your answer in terms of \(\mathbf{i}\) and \(\mathbf{j}\). [1 mark]
    • Question 13:
    AnswerMarks Guidance
    13(a)(i)Sums the forces given correctly AO1.1b
    1 2 3
    = 33i – 11j
    F  332 (11)2
    = 34.8 N (3 sf)
    Uses Pythagoras to find the
    magnitude of the vector and
    obtains correct magnitude
    AnswerMarks Guidance
    (given to 3 sig figs)AO1.1b B1
    (a)(ii)Uses trig expression with
    appropriate valuesAO1.1a M1
    tan
    33
    1
    θ = tan–1  
    3
    = 18.4 (3 sf)
    OR
    11
    sin
    1210
     11 
    θ = sin1
     
     1210
    = 18.4 (3 sf)
    OR
    33
    cos
    1210
     33 
    θ = cos1
     
     1210
    = 18.4 (3 sf)
    Obtains correct angle (given to
    AnswerMarks Guidance
    nearest 0.1)AO1.1b A1
    (b)States negative of ‘their’ part
    (a)(i)AO2.2a B1F
    4
    AnswerMarks Guidance
    Total5
    QMarking Instructions AO 
    Question 13:
--- 13(a)(i) ---
13(a)(i) | Sums the forces given correctly | AO1.1b | B1 | F = F + F + F
1 2 3
= 33i – 11j
F  332 (11)2
= 34.8 N (3 sf)
Uses Pythagoras to find the
magnitude of the vector and
obtains correct magnitude
(given to 3 sig figs) | AO1.1b | B1
(a)(ii) | Uses trig expression with
appropriate values | AO1.1a | M1 | 11
tan
33
1
θ = tan–1  
3
= 18.4 (3 sf)
OR
11
sin
1210
 11 
θ = sin1
 
 1210
= 18.4 (3 sf)
OR
33
cos
1210
 33 
θ = cos1
 
 1210
= 18.4 (3 sf)
Obtains correct angle (given to
nearest 0.1) | AO1.1b | A1
(b) | States negative of ‘their’ part
(a)(i) | AO2.2a | B1F | F = –33i + 11j
4
Total | 5
Q | Marking Instructions | AO | Marks | Typical Solution
    The three forces $\mathbf{F_1}$, $\mathbf{F_2}$ and $\mathbf{F_3}$ are acting on a particle.

$\mathbf{F_1} = (25\mathbf{i} + 12\mathbf{j})$ N

$\mathbf{F_2} = (-7\mathbf{i} + 5\mathbf{j})$ N

$\mathbf{F_3} = (15\mathbf{i} - 28\mathbf{j})$ N

The unit vectors $\mathbf{i}$ and $\mathbf{j}$ are horizontal and vertical respectively.

The resultant of these three forces is $\mathbf{F}$ newtons.

\begin{enumerate}[label=(\alph*)]
\begin{enumerate}[label=(\roman*)]
\item Find the magnitude of $\mathbf{F}$, giving your answer to three significant figures.
[2 marks]

\item Find the acute angle that $\mathbf{F}$ makes with the horizontal, giving your answer to the nearest 0.1°
[2 marks]
\end{enumerate}

\item The fourth force, $\mathbf{F_4}$, is applied to the particle so that the four forces are in equilibrium.

Find $\mathbf{F_4}$, giving your answer in terms of $\mathbf{i}$ and $\mathbf{j}$.
[1 mark]
\end{enumerate}

\hfill \mbox{\textit{AQA Paper 2  Q13 [5]}}
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