Edexcel M2 2003 January — Question 1 7 marks

Exam BoardEdexcel
ModuleM2 (Mechanics 2)
Year2003
SessionJanuary
Marks7
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCentre of Mass 1
TypeParticles at coordinate positions
DifficultyModerate -0.8 This is a straightforward centre of mass calculation requiring direct application of the standard formula. Part (a) uses the x-coordinate equation to find λ (simple algebra), and part (b) substitutes to find k using the y-coordinate equation. It's routine bookwork with no problem-solving insight needed, making it easier than average but not trivial since it requires careful algebraic manipulation across two parts.
Spec6.04b Find centre of mass: using symmetry

Three particles of mass \(3m\), \(5m\) and \(\lambda m\) are placed at points with coordinates \((4, 0)\), \((0, -3)\) and \((4, 2)\) respectively. The centre of mass of the system of three particles is at \((2, k)\).
  1. Show that \(\lambda = 2\). [4]
  2. Calculate the value of \(k\). [3]

Three particles of mass $3m$, $5m$ and $\lambda m$ are placed at points with coordinates $(4, 0)$, $(0, -3)$ and $(4, 2)$ respectively. The centre of mass of the system of three particles is at $(2, k)$.

\begin{enumerate}[label=(\alph*)]
\item Show that $\lambda = 2$. [4]
\item Calculate the value of $k$. [3]
\end{enumerate}

\hfill \mbox{\textit{Edexcel M2 2003 Q1 [7]}}