| Exam Board | Edexcel |
|---|---|
| Module | M1 (Mechanics 1) |
| Year | 2022 |
| Session | January |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Beam on point of tilting |
| Difficulty | Moderate -0.3 This is a standard M1 moments question requiring taking moments about the pivot point D when the beam is on the point of tilting. Part (a) involves one equilibrium equation, part (b) requires finding a threshold mass, and part (c) is a routine modelling statement. The calculations are straightforward with no geometric complexity or novel insight required, making it slightly easier than average for A-level. |
| Spec | 3.04b Equilibrium: zero resultant moment and force3.04c Use moments: beams, ladders, static problems |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| M(D): \(mg \times 1.2 = 30g \times 0.8\) | M1 A1 | Complete method for equation in \(m\) only; each equation must have correct no. of terms and be dimensionally correct |
| Other equations: \((\uparrow)\ R = mg + 30g\); M(A): \(2.5mg + 30g \times 4.5 = 3.7R\); M(G): \(30g \times 2 = 1.2R\); M(C): \(mg \times 2 = 0.8R\); M(B): \(2.5mg + 30g \times 0.5 = 1.3R\) | M0 if reaction at \(D\) not included | |
| \(m = 20\) (kg) | A1 | Allow inequality if \(m = 20\) stated at end |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| M(D): \(Xg \times 3.7 + 20g \times 1.2 = 30g \times 1.3\) | M1 A1ft | Complete method for equation in \(X\) only; follow through on their 20; allow inequality \(\geq\) |
| Other equations: \((\uparrow)\ S = mg + 30g + Xg\); M(A): \(2.5mg + 30g \times 5 = 3.7S\); M(G): \(30g \times 2.5 = 1.2S + Xg \times 2.5\); M(B): \(2.5mg + Xg \times 5 = 1.3S\) | ||
| \(X = \frac{150}{37}\), 4.1 or better (4.05405...) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| The mass of the block is concentrated at a point. | B1 | Must mention either mass or weight and 'acting at a point' or 'concentrated at a point' |
## Question 3:
**Part (a):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| M(D): $mg \times 1.2 = 30g \times 0.8$ | M1 A1 | Complete method for equation in $m$ only; each equation must have correct no. of terms and be dimensionally correct |
| Other equations: $(\uparrow)\ R = mg + 30g$; M(A): $2.5mg + 30g \times 4.5 = 3.7R$; M(G): $30g \times 2 = 1.2R$; M(C): $mg \times 2 = 0.8R$; M(B): $2.5mg + 30g \times 0.5 = 1.3R$ | | M0 if reaction at $D$ not included |
| $m = 20$ (kg) | A1 | Allow inequality if $m = 20$ stated at end |
**Part (b):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| M(D): $Xg \times 3.7 + 20g \times 1.2 = 30g \times 1.3$ | M1 A1ft | Complete method for equation in $X$ only; follow through on their 20; allow inequality $\geq$ |
| Other equations: $(\uparrow)\ S = mg + 30g + Xg$; M(A): $2.5mg + 30g \times 5 = 3.7S$; M(G): $30g \times 2.5 = 1.2S + Xg \times 2.5$; M(B): $2.5mg + Xg \times 5 = 1.3S$ | | |
| $X = \frac{150}{37}$, 4.1 or better (4.05405...) | A1 | cao |
**Part (c):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| The mass of the block is concentrated at a point. | B1 | Must mention either mass or weight and 'acting at a point' or 'concentrated at a point' |
---3.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f1bdc84b-c8a1-4e7c-a2ba-48b40c6a6d36-06_328_1356_244_296}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A beam $A D C B$ has length 5 m . The beam lies on a horizontal step with the end $A$ on the step and the end $B$ projecting over the edge of the step. The edge of the step is at the point $D$ where $D B = 1.3 \mathrm {~m}$, as shown in Figure 2.
When a small boy of mass 30 kg stands on the beam at $C$, where $C B = 0.5 \mathrm {~m}$, the beam is on the point of tilting.
The boy is modelled as a particle and the beam is modelled as a uniform rod.
\begin{enumerate}[label=(\alph*)]
\item Find the mass of the beam.
A block of mass $X \mathrm {~kg}$ is now placed on the beam at $A$.\\
The block is modelled as a particle.
\item Find the smallest value of $X$ that will enable the boy to stand on the beam at $B$ without the beam tilting.
\item State how you have used the modelling assumption that the block is a particle in your calculations.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M1 2022 Q3 [7]}}