OCR MEI Further Mechanics Minor 2019 June — Question 3 7 marks

Exam BoardOCR MEI
ModuleFurther Mechanics Minor (Further Mechanics Minor)
Year2019
SessionJune
Marks7
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Mark schemeDownload PDF ↗
TopicCentre of Mass 1
TypeL-shaped or composite rectangular lamina
DifficultyChallenging +1.8 This question requires students to find the centre of mass of an L-shaped composite lamina using standard formulas, then determine when this point lies outside the boundary by analyzing geometric constraints. While the setup is routine, the novel aspectβ€”finding when the COM lies *outside* the shapeβ€”requires careful geometric reasoning about the boundary conditions and solving an inequality involving the parameter k, making it significantly harder than standard COM calculations.
Spec6.04b Find centre of mass: using symmetry6.04c Composite bodies: centre of mass
3 Two identical uniform rectangular laminas, P and Q , each having length \(k a\) and width \(a\) are fixed together, in the same plane, to form a lamina R.
With reference to coordinate axes, the corners of P are at ( 0,0 ), ( \(k a , 0\) ), ( \(k a , a\) ) and ( \(0 , a\) ) and the corners of Q are at \(( k a , 0 ) , ( k a + a , 0 ) , ( k a + a , k a )\) and \(( k a , k a )\), as shown in Fig. 3. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{3b808042-95b8-4862-8355-3979c1981089-3_704_1102_459_244} \captionsetup{labelformat=empty} \caption{Fig. 3}
\end{figure} Determine the range of values of \(k\) for which the centre of mass of R lies outside the boundary of R.
Question 3:
AnswerMarks
3π‘˜π‘Ž π‘Ž
2π‘˜π‘₯Μ… = π‘˜Γ— +π‘˜ Γ—(π‘˜π‘Ž + )
AnswerMarks Guidance
2 2M1 3.1a
3 1
π‘₯Μ… [= π‘˜π‘Ž + π‘Ž] < π‘˜π‘Ž
AnswerMarks Guidance
4 4M1 1.1
π‘˜ > 1A1 2.1
π‘Ž π‘˜π‘Ž
2π‘˜π‘¦Μ… = π‘˜ Γ— +π‘˜Γ—
AnswerMarks Guidance
2 2M1 3.1a
1 1
𝑦̅ [= π‘˜π‘Ž+ π‘Ž] > π‘Ž
AnswerMarks Guidance
4 4M1 1.1
π‘˜ > 3A1 1.1
So π‘˜ > 3 onlyA1 2.2a
gained
[7]
Question 3:
3 | π‘˜π‘Ž π‘Ž
2π‘˜π‘₯Μ… = π‘˜Γ— +π‘˜ Γ—(π‘˜π‘Ž + )
2 2 | M1 | 3.1a
3 1
π‘₯Μ… [= π‘˜π‘Ž + π‘Ž] < π‘˜π‘Ž
4 4 | M1 | 1.1
π‘˜ > 1 | A1 | 2.1
π‘Ž π‘˜π‘Ž
2π‘˜π‘¦Μ… = π‘˜ Γ— +π‘˜Γ—
2 2 | M1 | 3.1a
1 1
𝑦̅ [= π‘˜π‘Ž+ π‘Ž] > π‘Ž
4 4 | M1 | 1.1
π‘˜ > 3 | A1 | 1.1
So π‘˜ > 3 only | A1 | 2.2a | Dep on all other marks being
gained
[7]
3 Two identical uniform rectangular laminas, P and Q , each having length $k a$ and width $a$ are fixed together, in the same plane, to form a lamina R.\\
With reference to coordinate axes, the corners of P are at ( 0,0 ), ( $k a , 0$ ), ( $k a , a$ ) and ( $0 , a$ ) and the corners of Q are at $( k a , 0 ) , ( k a + a , 0 ) , ( k a + a , k a )$ and $( k a , k a )$, as shown in Fig. 3.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{3b808042-95b8-4862-8355-3979c1981089-3_704_1102_459_244}
\captionsetup{labelformat=empty}
\caption{Fig. 3}
\end{center}
\end{figure}

Determine the range of values of $k$ for which the centre of mass of R lies outside the boundary of R.

\hfill \mbox{\textit{OCR MEI Further Mechanics Minor 2019 Q3 [7]}}
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