| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2018 |
| Session | Specimen |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Maximum or minimum mass |
| Difficulty | Moderate -0.3 This is a standard M1 moments question requiring taking moments about a point to find reaction forces, then using an inequality to find maximum mass. The setup is straightforward with clear geometry, and the methods are routine textbook applications of equilibrium conditions. Slightly easier than average due to the direct application of standard techniques with no conceptual surprises. |
| Spec | 3.04a Calculate moments: about a point3.04b Equilibrium: zero resultant moment and force |
| VIAV SIHI NI BIIIM ION OC | VGHV SIHI NI GHIYM ION OC | VJ4V SIHI NI JIIYM ION OC |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(M(A)\; (30g \times 2) + (50g \times 4) = 0.6S\) | M1 | Moments equation. Requires all terms dimensionally correct. Condone sign errors. Allow M1 if \(g\) missing |
| \(M(C)\; (0.6 \times R) = (1.4 \times 30g) + (3.4 \times 50g)\) or \(M(G)\; (2\times R) = (1.4 \times S) + (2 \times 50g)\) or \(M(B)\; (4\times R) + (2\times 30g) = (3.4 \times S)\) | A1 | Correct unsimplified equation |
| \((\uparrow)\; R + 30g + 50g = S \quad (R + 784 = S)\) | M1 | Resolve vertically. Requires all 4 terms. Condone sign errors |
| Correct equation (with \(R\) or their \(R\)) | A1 | |
| \(R = 3460\) or \(3500\) or \(\frac{1060g}{3}\) (N). Not \(353.3g\) | A1 | One force correct |
| \(S = 4250\) or \(4200\) or \(\frac{1300g}{3}\) (N). Not \(433.3g\) | A1 | Both forces correct. If both given as decimal multiples of \(g\), mark as accuracy penalty A0A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(M(C)\; (30g \times 1.4) + (Mg \times 3.4) = 0.6 \times 5000\) | M1 | Use \(R = 5000\) and complete method to form equation in \(M\) or weight. Needs all terms present and dimensionally correct. Condone sign errors. Accept inequality. Use of \(R\) and \(S\) from (a) is M0 |
| Correct equation in \(M\) (not weight) (implied by \(M = 77.68\)) | A1 | |
| \(M = 77\) kg | A1 | 77.7 is A0 even if penalty for over-specified answers already applied |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| The weight of the diver acts at a point. | B1 | Accept "the mass of the diver is at a point" |
## Question 4:
### Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $M(A)\; (30g \times 2) + (50g \times 4) = 0.6S$ | M1 | Moments equation. Requires all terms dimensionally correct. Condone sign errors. Allow M1 if $g$ missing |
| $M(C)\; (0.6 \times R) = (1.4 \times 30g) + (3.4 \times 50g)$ or $M(G)\; (2\times R) = (1.4 \times S) + (2 \times 50g)$ or $M(B)\; (4\times R) + (2\times 30g) = (3.4 \times S)$ | A1 | Correct unsimplified equation |
| $(\uparrow)\; R + 30g + 50g = S \quad (R + 784 = S)$ | M1 | Resolve vertically. Requires all 4 terms. Condone sign errors |
| Correct equation (with $R$ or their $R$) | A1 | |
| $R = 3460$ or $3500$ or $\frac{1060g}{3}$ (N). Not $353.3g$ | A1 | One force correct |
| $S = 4250$ or $4200$ or $\frac{1300g}{3}$ (N). Not $433.3g$ | A1 | Both forces correct. If both given as decimal multiples of $g$, mark as accuracy penalty A0A1 |
### Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $M(C)\; (30g \times 1.4) + (Mg \times 3.4) = 0.6 \times 5000$ | M1 | Use $R = 5000$ and complete method to form equation in $M$ or weight. Needs all terms present and dimensionally correct. Condone sign errors. Accept inequality. Use of $R$ and $S$ from (a) is M0 |
| Correct equation in $M$ (not weight) (implied by $M = 77.68$) | A1 | |
| $M = 77$ kg | A1 | 77.7 is A0 even if penalty for over-specified answers already applied |
### Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| The weight of the diver acts at a point. | B1 | Accept "the mass of the diver is at a point" |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{6ab8838f-d6f8-4761-8def-1022d97d4e82-10_238_1161_267_388}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A diving board $A B$ consists of a wooden plank of length 4 m and mass 30 kg . The plank is held at rest in a horizontal position by two supports at the points $A$ and $C$, where $A C = 0.6 \mathrm {~m}$, as shown in Figure 1. The force on the plank at $A$ acts vertically downwards and the force on the plank at $C$ acts vertically upwards.
A diver of mass 50 kg is standing on the board at the end $B$. The diver is modelled as a particle and the plank is modelled as a uniform rod. The plank is in equilibrium.
\begin{enumerate}[label=(\alph*)]
\item Find
\begin{enumerate}[label=(\roman*)]
\item the magnitude of the force acting on the plank at $A$,
\item the magnitude of the force acting on the plank at $C$.
The support at $A$ will break if subjected to a force whose magnitude is greater than 5000 N .
\end{enumerate}\item Find, in kg, the greatest integer mass of a diver who can stand on the board at $B$ without breaking the support at $A$.
\item Explain how you have used the fact that the diver is modelled as a particle.\\
\begin{center}
\begin{tabular}{|l|l|l|}
\hline
VIAV SIHI NI BIIIM ION OC & VGHV SIHI NI GHIYM ION OC & VJ4V SIHI NI JIIYM ION OC \\
\hline
\end{tabular}
\end{center}
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2018 Q4 [10]}}