| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2022 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Lamina with attached triangle |
| Difficulty | Standard +0.3 This is a standard M2 centre of mass question requiring systematic calculation of individual centres of mass, then combining using the formula. Part (a) involves routine arithmetic with given masses and coordinates. Part (b) applies the standard suspension equilibrium principle (tan θ = horizontal distance / vertical distance). While multi-step, it follows a well-practiced algorithm with no novel insight required, making it slightly easier than average. |
| Spec | 6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
7.
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\caption{Figure 3}
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The template shown in Figure 3 is formed by joining together three separate laminas. All three laminas lie in the same plane.
\begin{itemize}
\item PQUV is a uniform square lamina with sides of length $3 a$
\item URST is a uniform square lamina with sides of length $6 a$
\item $Q R U$ is a uniform triangular lamina with $U Q = 3 a , U R = 6 a$ and angle $Q U R = 90 ^ { \circ }$
\end{itemize}
The mass per unit area of $P Q U V$ is $k$, where $k$ is a constant.\\
The mass per unit area of URST is $k$.\\
The mass per unit area of $Q R U$ is $2 k$.\\
The distance of the centre of mass of the template from $Q T$ is $d$.
\begin{enumerate}[label=(\alph*)]
\item Show that $d = \frac { 29 } { 14 } a$
The template is freely suspended from the point $Q$ and hangs in equilibrium with $Q R$ at $\theta ^ { \circ }$ to the downward vertical.
\item Find the value of $\theta$\\
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\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2022 Q7 [12]}}