CAIE M2 2019 June — Question 3 5 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2019
SessionJune
Marks5
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Mark schemeDownload PDF ↗
TopicCentre of Mass 1
TypeSolid with removed cylinder or hemisphere from solid
DifficultyStandard +0.3 This is a standard centre of mass problem requiring subtraction of a hemisphere from a cylinder. Students apply the formula for composite bodies with given volume formulas, involving straightforward calculation of volumes, individual centres of mass, and the composite centre of mass formula. Slightly above average difficulty due to the 3D geometry and careful coordinate setup, but follows a well-practiced method with no novel insight required.
Spec6.04d Integration: for centre of mass of laminas/solids
3 \includegraphics[max width=\textwidth, alt={}, center]{111bcbf6-daaf-4d8d-9299-d591ac7369f1-05_448_802_258_676} The diagram shows the cross-section through the centre of mass of a uniform solid object. The object is a cylinder of radius 0.2 m and length 0.7 m , from which a hemisphere of radius 0.2 m has been removed at one end. The point \(A\) is the centre of the plane face at the other end of the object. Find the distance of the centre of mass of the object from \(A\).
[0pt] [The volume of a hemisphere is \(\frac { 2 } { 3 } \pi r ^ { 3 }\).]
Question 3:
AnswerMarks Guidance
AnswerMarks Guidance
Volume of cylinder \(= \pi \times 0.22 \times 0.7 (= 0.028\pi)\) AND Volume of hemisphere \(= 2\pi \times \frac{0.2^3}{3} (= 0.0053333\pi)\)B1 Both volumes required for B1
Distance of centre of mass from object base \(= 0.7 - 3 \times \frac{0.2}{8} (= 0.625)\)B1
\(x\left(\pi \times 0.2^2 \times 0.7 - 2\pi \times \frac{0.2^3}{3}\right) + \left(0.7 - 3 \times \frac{0.2}{8}\right) \times 2\pi \times \frac{0.2^3}{3} = 0.35 \times 0.028\pi\)M1A1 Take moments about the plane face
\(x = 0.285\) mA1 
## Question 3:

| Answer | Marks | Guidance |
|--------|-------|----------|
| Volume of cylinder $= \pi \times 0.22 \times 0.7 (= 0.028\pi)$ AND Volume of hemisphere $= 2\pi \times \frac{0.2^3}{3} (= 0.0053333\pi)$ | B1 | Both volumes required for B1 |
| Distance of centre of mass from object base $= 0.7 - 3 \times \frac{0.2}{8} (= 0.625)$ | B1 | |
| $x\left(\pi \times 0.2^2 \times 0.7 - 2\pi \times \frac{0.2^3}{3}\right) + \left(0.7 - 3 \times \frac{0.2}{8}\right) \times 2\pi \times \frac{0.2^3}{3} = 0.35 \times 0.028\pi$ | M1A1 | Take moments about the plane face |
| $x = 0.285$ m | A1 | |

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\includegraphics[max width=\textwidth, alt={}, center]{111bcbf6-daaf-4d8d-9299-d591ac7369f1-05_448_802_258_676}

The diagram shows the cross-section through the centre of mass of a uniform solid object. The object is a cylinder of radius 0.2 m and length 0.7 m , from which a hemisphere of radius 0.2 m has been removed at one end. The point $A$ is the centre of the plane face at the other end of the object. Find the distance of the centre of mass of the object from $A$.\\[0pt]
[The volume of a hemisphere is $\frac { 2 } { 3 } \pi r ^ { 3 }$.]\\

\hfill \mbox{\textit{CAIE M2 2019 Q3 [5]}}
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