| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2023 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Beam suspended by vertical ropes |
| Difficulty | Standard +0.3 This is a standard M1 moments question requiring taking moments about two points and solving simultaneous equations. The 'on the point of tilting' concept in part (b) is routine for this topic. While it has multiple parts, each step follows textbook procedures with no novel insight required, making it slightly easier than average. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(T\) and \(4T\) correctly placed | B1 | Correct relationship between tensions, seen or implied |
| Vertical resolution: \(T + 4T = pmg + mg\) OR a moments equation | M1 A1 | Vertical resolution. Condone forces at \(C\) and \(D\) the wrong way round or written as \(T_C\) and \(T_D\). May be replaced with a moments equation |
| \(\text{M}(A)\): \((4T \times 0.6) + (T \times 1.8) = (mg \times 1)\) | M1 A1 | Moments equation about a valid point with correct terms |
| \(5\left(\frac{5mg}{21}\right) = pmg + mg\) | M1 | Eliminate \(T\) |
| \(p = \frac{4}{21}\) (exact ratio of 2 positive integers) | A1 | Cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Tension at \(D\) is zero, seen or implied | B1 | |
| \(\text{M}(C)\): \((qmg \times 0.6) = (mg \times 0.4)\) | M1 A1 | |
| \(q = \frac{2}{3}\) (exact ratio of 2 positive integers), accept \(0.666...\) or \(0.\dot{6}\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| The centre of mass (or gravity) of the beam is in the middle; the mass (weight) of the beam acts at the middle; mass at centre; centre of mass at the centre. | B1 | Penalise incorrect extras |
# Question 4:
## Part 4(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T$ and $4T$ correctly placed | B1 | Correct relationship between tensions, seen or implied |
| Vertical resolution: $T + 4T = pmg + mg$ **OR** a moments equation | M1 A1 | Vertical resolution. Condone forces at $C$ and $D$ the wrong way round or written as $T_C$ and $T_D$. May be replaced with a moments equation |
| $\text{M}(A)$: $(4T \times 0.6) + (T \times 1.8) = (mg \times 1)$ | M1 A1 | Moments equation about a valid point with correct terms |
| $5\left(\frac{5mg}{21}\right) = pmg + mg$ | M1 | Eliminate $T$ |
| $p = \frac{4}{21}$ (exact ratio of 2 positive integers) | A1 | Cao |
## Part 4(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Tension at $D$ is zero, seen or implied | B1 | |
| $\text{M}(C)$: $(qmg \times 0.6) = (mg \times 0.4)$ | M1 A1 | |
| $q = \frac{2}{3}$ (exact ratio of 2 positive integers), accept $0.666...$ or $0.\dot{6}$ | A1 | |
## Part 4(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| The centre of mass (or gravity) of the beam is in the middle; the mass (weight) of the beam acts at the middle; mass at centre; centre of mass at the centre. | B1 | Penalise incorrect extras |4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f2737a11-4a15-41e9-9f87-31a705a8948b-08_625_1488_246_287}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
Figure 1 shows a beam $A B$, of mass $m \mathrm {~kg}$ and length 2 m , suspended by two light vertical ropes.\\
The ropes are attached to the points $C$ and $D$ on the beam, where $A C = 0.6 \mathrm {~m}$ and $D B = 0.2 \mathrm {~m}$\\
The beam is in equilibrium in a horizontal position.\\
A particle of mass pmkg is attached to the beam at $A$ and the beam remains in equilibrium in a horizontal position.
The beam is modelled as a uniform rod.
\begin{enumerate}[label=(\alph*)]
\item Given that the tension in the rope attached at $C$ is four times the tension in the rope attached at $D$, use the model to find the exact value of $p$.
The particle of mass $p m \mathrm {~kg}$ at $A$ is removed and replaced by a particle of mass $q m \mathrm {~kg}$ at $A$.\\
The beam remains in equilibrium in a horizontal position but is now on the point of tilting.
\item Using the model, find the exact value of $q$
\item State how you have used the modelling assumption that the beam is uniform.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2023 Q4 [12]}}