Edexcel M2 2006 January — Question 5 12 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Year2006
SessionJanuary
Marks12
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCentre of Mass 1
TypeParticles at coordinate positions
DifficultyStandard +0.3 This is a straightforward M2 centre of mass question requiring standard formulas. Part (a) uses the basic weighted average formula for particles, (b) applies the same formula with the lamina's centre at the centroid, (c) is immediate once k is found, and (d) uses simple trigonometry with the vertical through the suspension point. All techniques are routine with no novel problem-solving required, making it slightly easier than average.
Spec6.04a Centre of mass: gravitational effect6.04c Composite bodies: centre of mass6.04e Rigid body equilibrium: coplanar forces

\includegraphics{figure_1} Figure 1 shows a triangular lamina \(ABC\). The coordinates of \(A\), \(B\) and \(C\) are \((0, 4)\), \((9, 0)\) and \((0, -4)\) respectively. Particles of mass \(4m\), \(6m\) and \(2m\) are attached at \(A\), \(B\) and \(C\) respectively.
  1. Calculate the coordinates of the centre of mass of the three particles, without the lamina. [4]
The lamina \(ABC\) is uniform and of mass \(km\). The centre of mass of the combined system consisting of the three particles and the lamina has coordinates \((4, \lambda)\).
  1. Show that \(k = 6\). [3]
  2. Calculate the value of \(\lambda\). [2]
The combined system is freely suspended from \(O\) and hangs at rest.
  1. Calculate, in degrees to one decimal place, the angle between \(AC\) and the vertical. [3]

(a)
AnswerMarks
\(12m\bar{x} = 6m \times 9\)M1
\(\bar{x} = 4\frac{1}{2}\)A1
\(12m\bar{y} = 16m - 8m\)M1
\(\bar{y} = \frac{2}{3}\)A1
(b)
AnswerMarks
\((12 + k)m \times 4 = 12m \times 4\frac{1}{2} + km \times 3\) ft their \(\bar{x}\)M1 A1ft
\(k = 6\)A1
(c)
AnswerMarks
\(18m \times \lambda = 12m \times \frac{2}{3}\) \(\Rightarrow\) \(\lambda = \frac{4}{9}\)M1 A1
(d)
AnswerMarks
\(\tan\theta = \frac{4}{9}\) \(\Rightarrow\) \(\theta \approx 83.7°\) ft their \(\lambda\), caoM1 A1ft A1
## (a)
$12m\bar{x} = 6m \times 9$ | M1 |
$\bar{x} = 4\frac{1}{2}$ | A1 |

$12m\bar{y} = 16m - 8m$ | M1 |
$\bar{y} = \frac{2}{3}$ | A1 |

## (b)
$(12 + k)m \times 4 = 12m \times 4\frac{1}{2} + km \times 3$ ft their $\bar{x}$ | M1 A1ft |
$k = 6$ | A1 |

## (c)
$18m \times \lambda = 12m \times \frac{2}{3}$ $\Rightarrow$ $\lambda = \frac{4}{9}$ | M1 A1 |

## (d)
$\tan\theta = \frac{4}{9}$ $\Rightarrow$ $\theta \approx 83.7°$ ft their $\lambda$, cao | M1 A1ft A1 |

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\includegraphics{figure_1}

Figure 1 shows a triangular lamina $ABC$. The coordinates of $A$, $B$ and $C$ are $(0, 4)$, $(9, 0)$ and $(0, -4)$ respectively. Particles of mass $4m$, $6m$ and $2m$ are attached at $A$, $B$ and $C$ respectively.

\begin{enumerate}[label=(\alph*)]
\item Calculate the coordinates of the centre of mass of the three particles, without the lamina. [4]
\end{enumerate}

The lamina $ABC$ is uniform and of mass $km$. The centre of mass of the combined system consisting of the three particles and the lamina has coordinates $(4, \lambda)$.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{1}
\item Show that $k = 6$. [3]
\item Calculate the value of $\lambda$. [2]
\end{enumerate}

The combined system is freely suspended from $O$ and hangs at rest.

\begin{enumerate}[label=(\alph*)]
\setcounter{enumi}{3}
\item Calculate, in degrees to one decimal place, the angle between $AC$ and the vertical. [3]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M2 2006 Q5 [12]}}