CAIE M2 2019 June — Question 2 4 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2019
SessionJune
Marks4
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Mark schemeDownload PDF ↗
TopicCentre of Mass 2
TypeLamina with hole removed
DifficultyStandard +0.3 This is a standard composite lamina problem requiring the removal of a square from a larger square. Students apply the formula for composite bodies (using negative mass for the removed section) and calculate the center of mass using symmetry and basic coordinate geometry. It's slightly easier than average as it involves only two simple shapes with clear symmetry, requiring straightforward arithmetic rather than complex problem-solving.
Spec6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids
2 \includegraphics[max width=\textwidth, alt={}, center]{f3a35846-075d-4e03-ba6b-82774ef0e4f8-04_442_554_260_794} A uniform lamina \(A B C E F G\) is formed from a square \(A B D G\) by removing a smaller square \(C D F E\) from one corner. \(A B = 0.7 \mathrm {~m}\) and \(D F = 0.3 \mathrm {~m}\) (see diagram). Find the distance of the centre of mass of the lamina from \(A\).
Question 2:
AnswerMarks Guidance
Original square: Area \(= 0.7^2\), CoM \(= \sqrt{(0.35^2 + 0.35^2)}\) and Smaller square: Area \(= 0.3^2\), CoM \(= \sqrt{(0.15^2 + 0.15^2)}\)B1 \(0.49, 0.495\) from A; \(0.09, 0.21213\ldots\) from D or E
\(AX(0.49 - 0.09) + 0.09\!\left(\sqrt{0.98} - \sqrt{0.045}\right) = 0.49 \times 0.495\)M1A1 Attempt to take moments about A
\(AX = 0.431 \text{ m}\)A1

Total: 4 marks

Alternative method for Question 2:
AnswerMarks Guidance
\((0.49 \times 0.35) = (0.09 \times 0.55) + 0.4X \rightarrow X = 0.305\)M1 Take moments about AG or AB
\(X = Y = 0.305\)B1
Question 2:
AnswerMarks Guidance
\(AX = \sqrt{(0.305^2 + 0.305^2)}\)M1 Use Pythagoras's theorem
\(AX = 0.431\)A1 
**Question 2:**

Original square: Area $= 0.7^2$, CoM $= \sqrt{(0.35^2 + 0.35^2)}$ and Smaller square: Area $= 0.3^2$, CoM $= \sqrt{(0.15^2 + 0.15^2)}$ | B1 | $0.49, 0.495$ from A; $0.09, 0.21213\ldots$ from D or E

$AX(0.49 - 0.09) + 0.09\!\left(\sqrt{0.98} - \sqrt{0.045}\right) = 0.49 \times 0.495$ | M1A1 | Attempt to take moments about A

$AX = 0.431 \text{ m}$ | A1 |

Total: 4 marks

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**Alternative method for Question 2:**

$(0.49 \times 0.35) = (0.09 \times 0.55) + 0.4X \rightarrow X = 0.305$ | M1 | Take moments about AG or AB
$X = Y = 0.305$ | B1 |

## Question 2:

| $AX = \sqrt{(0.305^2 + 0.305^2)}$ | M1 | Use Pythagoras's theorem |
| $AX = 0.431$ | A1 | |

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\includegraphics[max width=\textwidth, alt={}, center]{f3a35846-075d-4e03-ba6b-82774ef0e4f8-04_442_554_260_794}

A uniform lamina $A B C E F G$ is formed from a square $A B D G$ by removing a smaller square $C D F E$ from one corner. $A B = 0.7 \mathrm {~m}$ and $D F = 0.3 \mathrm {~m}$ (see diagram). Find the distance of the centre of mass of the lamina from $A$.\\

\hfill \mbox{\textit{CAIE M2 2019 Q2 [4]}}
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