CAIE M2 2010 November — Question 2 6 marks

Exam BoardCAIE
ModuleM2 (Mechanics 2)
Year2010
SessionNovember
Marks6
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCentre of Mass 1
TypeFrame with circular arc or semicircular arc components
DifficultyStandard +0.3 This is a straightforward centre of mass problem requiring students to find the centroid of a circular arc (standard formula), apply the centre of mass formula for a composite system, and solve a linear equation. While it involves multiple components and requires knowing or deriving the arc centroid formula, the problem follows a standard template with no novel insight required.
Spec6.04c Composite bodies: centre of mass
2 \includegraphics[max width=\textwidth, alt={}, center]{af7d1fc8-5552-48b8-a359-895b2b5d3d6c-2_673_401_525_872} A bow consists of a uniform curved portion \(A B\) of mass 1.4 kg , and a uniform taut string of mass \(m \mathrm {~kg}\) which joins \(A\) and \(B\). The curved portion \(A B\) is an arc of a circle centre \(O\) and radius 0.8 m . Angle \(A O B\) is \(\frac { 2 } { 3 } \pi\) radians (see diagram). The centre of mass of the bow (including the string) is 0.65 m from \(O\). Calculate \(m\).
Question 2:
AnswerMarks Guidance
Working/AnswerMark Guidance
\(OG = 0.8\sin(\pi/3)/(\pi/3)\)B1 0.66159
\(OM = 0.8\cos(\pi/3)\)B1 0.4
M1For taking moments about O
\(0.65(m + 1.4) = 0.4m + 0.66159 \times 1.4\)A1
\(0.25m = 0.01159 \times 1.4\)M1 For collecting like terms
\(m = 0.0649\)A1 [6]
OR:
AnswerMarks Guidance
Working/AnswerMark Guidance
\(OG = 0.8\sin(\pi/3)/(\pi/3)\)B1 0.66159
\(OM = 0.8\cos(\pi/3)\)B1 0.4
M1Taking moments about M
\((1.4 + m) \times 0.25 = 1.4 \times 0.26159\)A1
\(0.25m = 1.4 \times 0.01159\)M1 For collecting like terms
\(m = 0.0649\)A1 [6] 
## Question 2:

| Working/Answer | Mark | Guidance |
|---|---|---|
| $OG = 0.8\sin(\pi/3)/(\pi/3)$ | B1 | 0.66159 |
| $OM = 0.8\cos(\pi/3)$ | B1 | 0.4 |
| | M1 | For taking moments about O |
| $0.65(m + 1.4) = 0.4m + 0.66159 \times 1.4$ | A1 | |
| $0.25m = 0.01159 \times 1.4$ | M1 | For collecting like terms |
| $m = 0.0649$ | A1 [6] | |

**OR:**

| Working/Answer | Mark | Guidance |
|---|---|---|
| $OG = 0.8\sin(\pi/3)/(\pi/3)$ | B1 | 0.66159 |
| $OM = 0.8\cos(\pi/3)$ | B1 | 0.4 |
| | M1 | Taking moments about M |
| $(1.4 + m) \times 0.25 = 1.4 \times 0.26159$ | A1 | |
| $0.25m = 1.4 \times 0.01159$ | M1 | For collecting like terms |
| $m = 0.0649$ | A1 [6] | |

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2\\
\includegraphics[max width=\textwidth, alt={}, center]{af7d1fc8-5552-48b8-a359-895b2b5d3d6c-2_673_401_525_872}

A bow consists of a uniform curved portion $A B$ of mass 1.4 kg , and a uniform taut string of mass $m \mathrm {~kg}$ which joins $A$ and $B$. The curved portion $A B$ is an arc of a circle centre $O$ and radius 0.8 m . Angle $A O B$ is $\frac { 2 } { 3 } \pi$ radians (see diagram). The centre of mass of the bow (including the string) is 0.65 m from $O$. Calculate $m$.

\hfill \mbox{\textit{CAIE M2 2010 Q2 [6]}}
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