| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2024 |
| Session | June |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Maximum or minimum mass |
| Difficulty | Standard +0.3 This is a standard M1 moments problem requiring taking moments about a pivot point to find equilibrium conditions. Part (a) involves finding minimum mass by considering the limiting case (tilting about C), and part (b) requires calculating reaction force using vertical equilibrium. The setup is straightforward with clearly defined distances, requiring only basic moment principles and no novel insight—slightly easier than average A-level mechanics. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(M(C)\): \(Mg \times 4.5 + 1.2g \times 2 = 4g \times 1.5\) | M1A1 | M1: Equation in \(M\) only, correct number of terms, condone sign errors and missing \(g\)'s |
| \(M = 0.8\) oe | A1 (3) | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(M(E)\): \(R_C \times 0.6 = 1.2g \times 2.6\) | M1A1 | M1: Equation in \(R_C\) only, correct number of terms, condone sign errors and missing \(g\)'s |
| \(R_C = 5.2g\) isw | A1 (3) | \(\frac{26g}{5}\), 51 or 51.0 |
# Question 4:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $M(C)$: $Mg \times 4.5 + 1.2g \times 2 = 4g \times 1.5$ | M1A1 | M1: Equation in $M$ only, correct number of terms, condone sign errors and missing $g$'s |
| $M = 0.8$ oe | A1 (3) | cao |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $M(E)$: $R_C \times 0.6 = 1.2g \times 2.6$ | M1A1 | M1: Equation in $R_C$ only, correct number of terms, condone sign errors and missing $g$'s |
| $R_C = 5.2g$ isw | A1 (3) | $\frac{26g}{5}$, 51 or 51.0 |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{7a65555e-1bb2-4947-8e70-50f267017bfd-08_417_1745_378_258}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A non-uniform rod $A B$ has length 6.5 m and mass 1.2 kg . The centre of mass of the rod is 3 m from $A$. The rod rests on a horizontal step and overhangs the end of the step $C$ by 1.5 m , as shown in Figure 2.
The rod is perpendicular to the edge of the step.\\
A particle of mass 4 kg is placed on the rod at $B$ and another particle, whose mass is $M \mathrm {~kg}$, is placed on the rod at $D$, where $A D = 0.5 \mathrm {~m}$.
The rod remains in equilibrium in a horizontal position.
\begin{enumerate}[label=(\alph*)]
\item Find the smallest possible value of $M$.
The particle at $B$ and the particle at $D$ are now removed.\\
A new particle is placed on the rod at the point $E$, where $E B = 0.9 \mathrm {~m}$.\\
The rod remains in equilibrium in a horizontal position but is on the point of tilting about $C$.
\item Find the magnitude of the force acting on the rod at $C$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2024 Q4 [6]}}