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
TypeCentre of mass with variable parameter
DifficultyStandard +0.3 This is a straightforward centre of mass problem requiring knowledge of the standard result for the centroid of a circular arc (2r sin(θ)/θ from centre), then applying the composite centre of mass formula. It involves multiple steps but uses bookwork formulas with clear substitution, making it slightly easier than average.
Spec6.04c Composite bodies: centre of mass

\includegraphics{figure_2} A bow consists of a uniform curved portion \(AB\) of mass \(1.4 \text{ kg}\), and a uniform taut string of mass \(m \text{ kg}\) which joins \(A\) and \(B\). The curved portion \(AB\) is an arc of a circle centre \(O\) and radius \(0.8 \text{ m}\). Angle \(AOB\) is \(\frac{2}{3}\pi\) radians (see diagram). The centre of mass of the bow (including the string) is \(0.65 \text{ m}\) from \(O\). Calculate \(m\). [6]

Question 2:
AnswerMarks Guidance
2π π
OG = 0.8sin( /3)/( /3)B1 0.66159
π
AnswerMarks Guidance
OM = 0.8cos( /3)B1 0.4
M1For taking moments about O
0.65(m + 1.4) = 0.4m + 0.66159x1.4A1
0.25m = 0.01159 x 1.4M1 For collecting like terms
m = 0.0649A1
ORπ π
OG = 0.8sin( /3)/( /3)B1 0.66159
π
AnswerMarks Guidance
OM = 0.8cos( /3)B1 0.4
M1Taking moments about M
(1.4 + m) × 0.25 = 1.4 × 0.26159A1
0.25m = 1.4x0.01159M1 For collecting like terms
m = 0.0649A1
[6]
Question 2:
2 | π π
OG = 0.8sin( /3)/( /3) | B1 | 0.66159
π
OM = 0.8cos( /3) | B1 | 0.4
M1 | For taking moments about O
0.65(m + 1.4) = 0.4m + 0.66159x1.4 | A1
0.25m = 0.01159 x 1.4 | M1 | For collecting like terms
m = 0.0649 | A1
OR | π π
OG = 0.8sin( /3)/( /3) | B1 | 0.66159
π
OM = 0.8cos( /3) | B1 | 0.4
M1 | Taking moments about M
(1.4 + m) × 0.25 = 1.4 × 0.26159 | A1
0.25m = 1.4x0.01159 | M1 | For collecting like terms
m = 0.0649 | A1
[6]
\includegraphics{figure_2}

A bow consists of a uniform curved portion $AB$ of mass $1.4 \text{ kg}$, and a uniform taut string of mass $m \text{ kg}$ which joins $A$ and $B$. The curved portion $AB$ is an arc of a circle centre $O$ and radius $0.8 \text{ m}$. Angle $AOB$ is $\frac{2}{3}\pi$ radians (see diagram). The centre of mass of the bow (including the string) is $0.65 \text{ m}$ from $O$. Calculate $m$. [6]

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