Edexcel M2 2024 October — Question 4

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Year2024
SessionOctober
PaperDownload PDF ↗
TopicMoments
TypeCentre of mass of composite shapes
DifficultyStandard +0.3 This is a standard M2 centres of mass question requiring (a) finding the centroid of a trapezium using the standard formula or decomposition, and (b) applying the equilibrium condition that the centre of mass hangs directly below the suspension point. Both parts use routine techniques with straightforward calculations, making it slightly easier than average for M2.
Spec6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces
4. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{3e78f951-041d-4227-aa4b-e67a6ab5b4cd-10_301_871_319_598} \captionsetup{labelformat=empty} \caption{Figure 2}
\end{figure} The uniform lamina \(A B C D\) shown in Figure 2 is in the shape of an isosceles trapezium.
  • \(B C\) is parallel to \(A D\) and angle \(B A D\) is equal to angle \(A D C\)
  • \(B C = 5 a\) and \(A D = 7 a\)
  • the perpendicular distance between \(B C\) and \(A D\) is \(3 a\)
  • the distance of the centre of mass of \(A B C D\) from \(A D\) is \(d\)
    1. Show that \(d = \frac { 17 } { 12 } a\)
The uniform lamina \(P Q R S\) is a rectangle with \(P Q = 5 a\) and \(Q R = 9 a\).
The lamina \(A B C D\) in Figure 2 is used to cut a hole in \(P Q R S\) to form the template shown shaded in Figure 3. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{3e78f951-041d-4227-aa4b-e67a6ab5b4cd-10_364_876_1567_593} \captionsetup{labelformat=empty} \caption{Figure 3}
\end{figure}
The template is freely suspended from \(P\) and hangs in equilibrium with \(P S\) at an angle of \(\theta ^ { \circ }\) to the downward vertical.
  • Find the value of \(\theta\)
    • Question 4:
    AnswerMarks Guidance
    4(a)Rectangle 2 x Triangle
    B1
    AnswerMarks Guidance
    Mass1 5 a 2 3
    2 a2
    AnswerMarks
    21 8 a 2
    From
    AnswerMarks
    AD3
    a
    AnswerMarks Guidance
    2a d
    Some examples of alternatives
    AnswerMarks
    Moments about ADM1
    3
    1 5 a 2  a + 3 a 2  a = 1 8 a 2 d
    AnswerMarks
    2A1
    5 1 1 7 a
    a = 1 8 d  d = *
    AnswerMarks
    2 1 2A1*
    (5)
    4(a)
    AnswerMarks
    B1Correct mass ratio.
    B1Correct distances from AD, for an appropriate division of ABCD
    Distances may be measured from a parallel axis
    AnswerMarks
    M1Dimensionally consistent equation with all required terms. Accept
    working from a parallel axis. Accept an equation embedded in
    vector form.
    AnswerMarks
    A1Correct unsimplified equation. Accept embedded in vector form.
    A1*Obtain given answer from correct working. Working must include
    simplification or rearrangement (may be seen from table to
    equation). Answer must be extracted from vector form and have d =
    AnswerMarks
    4(b)9
    x = a
    AnswerMarks
    2B1
    Moments about PS or a parallel axis for LaminaM1
     17 
    From PS: 27a2y =45a22.5a−18a2 a+ a
     12 
     1 7 
    From AD: 2 7 a 2 Y = 4 5 a 2  1 .5 a − 1 8 a 2 a
    AnswerMarks
    1 2A1
    A1
    69 23
    y = a= a
    AnswerMarks
    27 9A1
    Expression for relevant angleM1
    y 4 6  
    t a n   = =
    ( )
    " 9 a " 8 1
    2
    AnswerMarks
    2 9 .6  =A1
    (7)
    (12)
    Notes
    4(b)
    AnswerMarks
    B1Distance of c of m of lamina from PQ seen or implied. May not be
    seen until the trig ratio.
    AnswerMarks
    M1Complete method to find c of m of Lamina (remaining mass) from
    PS or a parallel axis. Dimensionally consistent equation containing
    all required terms. If one distance is from PS and the other is from
    AD, treat as an accuracy error.
    AnswerMarks
    A1Unsimplified equation for PS or their parallel axis, with at most one
    error
    AnswerMarks
    A1Correct unsimplified equation for PS or their parallel axis.
    A1Correct distance of c of m from PS, 2.6aor better
    M1( )
    9 a
    Correct use of trig for a relevant angle where " " is their
    2
    distance of the c of m of the lamina from PQ and y is their
    calculated distance from PS.
    Allow with both a’s or neither. Allow reciprocal.
    AnswerMarks Guidance
    A130 or better (29.59229…) cao
    QuestionScheme Marks 
    Question 4:
--- 4(a) ---
4(a) | Rectangle | 2 x Triangle | Trapezium | B1
B1
Mass | 1 5 a 2 | 3
2 a2
2 | 1 8 a 2
From
AD | 3
a
2 | a | d
Some examples of alternatives
Moments about AD | M1
3
1 5 a 2  a + 3 a 2  a = 1 8 a 2 d
2 | A1
5 1 1 7 a
a = 1 8 d  d = *
2 1 2 | A1*
(5)
4(a)
B1 | Correct mass ratio.
B1 | Correct distances from AD, for an appropriate division of ABCD
Distances may be measured from a parallel axis
M1 | Dimensionally consistent equation with all required terms. Accept
working from a parallel axis. Accept an equation embedded in
vector form.
A1 | Correct unsimplified equation. Accept embedded in vector form.
A1* | Obtain given answer from correct working. Working must include
simplification or rearrangement (may be seen from table to
equation). Answer must be extracted from vector form and have d =
--- 4(b) ---
4(b) | 9
x = a
2 | B1
Moments about PS or a parallel axis for Lamina | M1
 17 
From PS: 27a2y =45a22.5a−18a2 a+ a
 12 
 1 7 
From AD: 2 7 a 2 Y = 4 5 a 2  1 .5 a − 1 8 a 2 a
1 2 | A1
A1
69 23
y = a= a
27 9 | A1
Expression for relevant angle | M1
y 4 6  
t a n   = =
( )
" 9 a " 8 1
2
2 9 .6  = | A1
(7)
(12)
Notes
4(b)
B1 | Distance of c of m of lamina from PQ seen or implied. May not be
seen until the trig ratio.
M1 | Complete method to find c of m of Lamina (remaining mass) from
PS or a parallel axis. Dimensionally consistent equation containing
all required terms. If one distance is from PS and the other is from
AD, treat as an accuracy error.
A1 | Unsimplified equation for PS or their parallel axis, with at most one
error
A1 | Correct unsimplified equation for PS or their parallel axis.
A1 | Correct distance of c of m from PS, 2.6aor better
M1 | ( )
9 a
Correct use of trig for a relevant angle where " " is their
2
distance of the c of m of the lamina from PQ and y is their
calculated distance from PS.
Allow with both a’s or neither. Allow reciprocal.
A1 | 30 or better (29.59229…) cao
Question | Scheme | Marks
    4.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{3e78f951-041d-4227-aa4b-e67a6ab5b4cd-10_301_871_319_598}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}

The uniform lamina $A B C D$ shown in Figure 2 is in the shape of an isosceles trapezium.

\begin{itemize}
  \item $B C$ is parallel to $A D$ and angle $B A D$ is equal to angle $A D C$
  \item $B C = 5 a$ and $A D = 7 a$
  \item the perpendicular distance between $B C$ and $A D$ is $3 a$
  \item the distance of the centre of mass of $A B C D$ from $A D$ is $d$
\begin{enumerate}[label=(\alph*)]
\item Show that $d = \frac { 17 } { 12 } a$
\end{itemize}

The uniform lamina $P Q R S$ is a rectangle with $P Q = 5 a$ and $Q R = 9 a$.\\
The lamina $A B C D$ in Figure 2 is used to cut a hole in $P Q R S$ to form the template shown shaded in Figure 3.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{3e78f951-041d-4227-aa4b-e67a6ab5b4cd-10_364_876_1567_593}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\end{center}
\end{figure}

\begin{itemize}
  \item $\quad P S$ is parallel to $A D$
  \item the perpendicular distance between $P S$ and $A D$ is a
  \item the perpendicular distance of $A$ from $P Q$ is $a$
\end{itemize}

The template is freely suspended from $P$ and hangs in equilibrium with $P S$ at an angle of $\theta ^ { \circ }$ to the downward vertical.
\item Find the value of $\theta$
\end{enumerate}

\hfill \mbox{\textit{Edexcel M2 2024 Q4}}
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