OCR Further Mechanics 2021 November — Question 8 13 marks

Exam BoardOCR
ModuleFurther Mechanics (Further Mechanics)
Year2021
SessionNovember
Marks13
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCentre of Mass 1
TypeToppling on inclined plane
DifficultyStandard +0.8 This is a multi-part centre of mass question requiring coordinate geometry, toppling conditions, and dynamic analysis. Parts (a)-(c) are standard applications of centre of mass formulas and toppling criteria, but parts (d)-(e) require tracking a moving particle and determining which edge causes toppling as the system's centre of mass shifts—this adds problem-solving complexity beyond routine exercises.
Spec6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces

8 A rectangular lamina of mass \(M\) has vertices at the origin \(O ( 0,0 ) , A ( 24 a , 0 ) , B ( 24 a , 6 a )\) and \(C ( 0,6 a )\), where \(a\) is a positive constant. A small object \(P\) of mass \(m\) is attached to the lamina at the point ( \(x , y\) ). The centre of mass of the system consisting of the lamina and \(P\) is at the point ( \(\mathrm { x } , \mathrm { y }\) ). \(P\) is modelled as a particle and the lamina is modelled as being uniform.
  1. Show that \(x = \frac { 12 M a + m x } { M + m }\).
  2. Find a corresponding expression for \(\bar { y }\). The lamina, with \(P\) no longer attached, is placed on a horizontal rectangular table, with its sides parallel to the edges of the table, and partly overhanging the edges of the table, as shown in the diagram. The corner of the table is at the point ( \(6 a , 2 a\) ). \includegraphics[max width=\textwidth, alt={}, center]{c6445493-9802-46ca-b7eb-7738a831d9ee-6_538_1431_849_246} When \(P\) is placed on the lamina at \(O\), the lamina topples over one of the edges of the table.
  3. Show that \(\mathrm { m } > \frac { 1 } { 2 } \mathrm { M }\). The lamina is now put back on the table in the same position as before. \(P\) is placed at the point \(( 12 a , 6 a )\) on the smooth upper surface of the lamina, and is projected towards \(O\). At a subsequent instant during the motion, \(P\) is at the point (12ak, 6ak) where \(0 < k < 1\).
  4. Assuming that the lamina has not yet toppled, find, in terms of \(M\) and \(m\), the value of \(k\) for which the centre of mass of the system lies on the table edge parallel to \(O C\).
  5. For the case \(\mathrm { m } = \frac { 3 } { 2 } \mathrm { M }\), determine which table edge the lamina topples over.

Question 8:
AnswerMarks Guidance
8(a) 12a×M +x×m 12Ma+mx
x = =
AnswerMarks Guidance
M +m M +mB1
[1]1.1 AG. www
8(b) 3a×M + y×m 3Ma+my
y = =
AnswerMarks Guidance
M +m M +mB1
[1]1.1
8(c) 12Ma 3Ma
If P is at O, x = and y =
M +m M +m
y <2a⇒3M <2M +2m⇒m> 1M
2
x <6a⇒12M <6M +6m⇒m>M
Conclusion: m> 1M
AnswerMarks
2B1ft
M1
M1
A1
AnswerMarks
[4]3.3
3.4
3.4
AnswerMarks
2.4FT their expression for
𝑦𝑦�
AnswerMarks
AG.Alternative:
B1 for correct expressions for
M1: forming 2 inequalities with 2a
and 6a (must be right way arou𝑥𝑥n̅,d𝑦𝑦�)
M1: simplifying or manipulating
both inequalities so that they can be
combined or compared
A1: fully correct and conclusion
www
AnswerMarks Guidance
8(d) 12Ma+m×12ak
x = used
M +m
12Ma+m×12ak
=6a
M +m
m−M
k = oe
AnswerMarks
2mB1
M1
A1
AnswerMarks
[3]3.3
3.4
AnswerMarks
1.1Their x equated to 6a
1( M )
k = 1−
AnswerMarks
2 mIgnore working with y
Ignore working with y unless this
affects final answer
AnswerMarks Guidance
8(e) 3 1
m= M ⇒k =
OC
2 6
3
3Ma+ M×6ak
y = 2
3
M + M
AnswerMarks
2B1
M13.3
3.43 
k = =0.16
OC
18
Substituting y = 6ak and
3
m= M into their y
2
6a+18ak 2
y =2a⇒ =2a⇒k =
OA
5 9
(k changes from 1 to 0 and k >k so)
OA OC
AnswerMarks
lamina topples over edge OA.A1
A1
AnswerMarks
[4]3.4
2.2a4 
k = =0.2
OA
18
www
PMT
OCR (Oxford Cambridge and RSA Examinations)
The Triangle Building
Shaftesbury Road
Cambridge
CB2 8EA
OCR Customer Contact Centre
Education and Learning
Telephone: 01223 553998
Facsimile: 01223 552627
Email: general.qualifications@ocr.org.uk
www.ocr.org.uk
For staff training purposes and as part of our quality assurance programme your call may be
recorded or monitored
Question 8:
8 | (a) | 12a×M +x×m 12Ma+mx
x = =
M +m M +m | B1
[1] | 1.1 | AG. www
8 | (b) | 3a×M + y×m 3Ma+my
y = =
M +m M +m | B1
[1] | 1.1
8 | (c) | 12Ma 3Ma
If P is at O, x = and y =
M +m M +m
y <2a⇒3M <2M +2m⇒m> 1M
2
x <6a⇒12M <6M +6m⇒m>M
Conclusion: m> 1M
2 | B1ft
M1
M1
A1
[4] | 3.3
3.4
3.4
2.4 | FT their expression for
𝑦𝑦�
AG. | Alternative:
B1 for correct expressions for
M1: forming 2 inequalities with 2a
and 6a (must be right way arou𝑥𝑥n̅,d𝑦𝑦�)
M1: simplifying or manipulating
both inequalities so that they can be
combined or compared
A1: fully correct and conclusion
www
8 | (d) | 12Ma+m×12ak
x = used
M +m
12Ma+m×12ak
=6a
M +m
m−M
k = oe
2m | B1
M1
A1
[3] | 3.3
3.4
1.1 | Their x equated to 6a
1( M )
k = 1−
2 m | Ignore working with y
Ignore working with y unless this
affects final answer
8 | (e) | 3 1
m= M ⇒k =
OC
2 6
3
3Ma+ M×6ak
y = 2
3
M + M
2 | B1
M1 | 3.3
3.4 | 3 
k = =0.16
OC
18
Substituting y = 6ak and
3
m= M into their y
2
6a+18ak 2
y =2a⇒ =2a⇒k =
OA
5 9
(k changes from 1 to 0 and k >k so)
OA OC
lamina topples over edge OA. | A1
A1
[4] | 3.4
2.2a | 4 
k = =0.2
OA
18
www
PMT
OCR (Oxford Cambridge and RSA Examinations)
The Triangle Building
Shaftesbury Road
Cambridge
CB2 8EA
OCR Customer Contact Centre
Education and Learning
Telephone: 01223 553998
Facsimile: 01223 552627
Email: general.qualifications@ocr.org.uk
www.ocr.org.uk
For staff training purposes and as part of our quality assurance programme your call may be
recorded or monitored
8 A rectangular lamina of mass $M$ has vertices at the origin $O ( 0,0 ) , A ( 24 a , 0 ) , B ( 24 a , 6 a )$ and $C ( 0,6 a )$, where $a$ is a positive constant. A small object $P$ of mass $m$ is attached to the lamina at the point ( $x , y$ ). The centre of mass of the system consisting of the lamina and $P$ is at the point ( $\mathrm { x } , \mathrm { y }$ ). $P$ is modelled as a particle and the lamina is modelled as being uniform.
\begin{enumerate}[label=(\alph*)]
\item Show that $x = \frac { 12 M a + m x } { M + m }$.
\item Find a corresponding expression for $\bar { y }$.

The lamina, with $P$ no longer attached, is placed on a horizontal rectangular table, with its sides parallel to the edges of the table, and partly overhanging the edges of the table, as shown in the diagram. The corner of the table is at the point ( $6 a , 2 a$ ).\\
\includegraphics[max width=\textwidth, alt={}, center]{c6445493-9802-46ca-b7eb-7738a831d9ee-6_538_1431_849_246}

When $P$ is placed on the lamina at $O$, the lamina topples over one of the edges of the table.
\item Show that $\mathrm { m } > \frac { 1 } { 2 } \mathrm { M }$.

The lamina is now put back on the table in the same position as before. $P$ is placed at the point $( 12 a , 6 a )$ on the smooth upper surface of the lamina, and is projected towards $O$. At a subsequent instant during the motion, $P$ is at the point (12ak, 6ak) where $0 < k < 1$.
\item Assuming that the lamina has not yet toppled, find, in terms of $M$ and $m$, the value of $k$ for which the centre of mass of the system lies on the table edge parallel to $O C$.
\item For the case $\mathrm { m } = \frac { 3 } { 2 } \mathrm { M }$, determine which table edge the lamina topples over.
\end{enumerate}

\hfill \mbox{\textit{OCR Further Mechanics 2021 Q8 [13]}}