| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2021 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Framework or multiple rod structures |
| Difficulty | Challenging +1.2 This is a multi-step moments problem requiring students to find the center of mass of a composite rigid body made from three uniform rods, then apply equilibrium conditions. While it involves several calculations (finding individual centers of mass, combining them, and using geometry), the techniques are standard M2 content with no novel insights required. The geometric setup is clearly defined, making it more straightforward than problems requiring students to deduce the configuration themselves. |
| Spec | 6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids |
| VIIV SIHI NI III HM ION OC | VIUV SIHI NI JIHM I ON OO | VI4V SIHI NI JIIIM ION OO |
| Answer | Marks | Guidance |
|---|---|---|
| \(6m \times 3a + 4m \times 4a + 5m \times 2a = 15m \times y\) | M1 | Moments about horizontal axis. Terms dimensionally consistent. Condone slip with \(a\). Needs all terms. Condone sign errors. |
| \((44ma = 15my)\) | A1 | Correct unsimplified |
| \(y = \frac{44a}{15}\) from \(B\) | A1 | Or equivalent. Correct for their axis: \(\frac{46a}{15}\) from \(A\), \(\frac{16a}{15}\) from \(E\) (CD) |
| Answer | Marks | Guidance |
|---|---|---|
| \(5m \times \frac{3a}{2} + 4m \times a = 15mx\) | M1 | Moments about vertical axis. Terms dimensionally consistent. Condone slip with \(a\). Needs all terms. Condone sign errors. |
| \(\left(\frac{23ma}{2} = 15mx\right)\) | A1 | Correct unsimplified |
| \(x = \frac{23a}{30}\) from \(E\) (\(AB\)) | A1 | Or equivalent. Correct for their axis: \(\frac{67a}{30}\) from \(C\) |
| Answer | Marks | Guidance |
|---|---|---|
| \(\tan\theta = \dfrac{\frac{23}{30}}{\frac{44}{15}}\) | M1 | Find a relevant angle using distances measured from \(B\). Allow for \(\tan\theta = \frac{88}{23}\) |
| \(= \frac{23}{88}(= 0.261...)\) | A1ft | Correct for their distances from \(B\). \(\binom{\text{horizontal}}{\text{vertical}}\) |
| \(\theta = 14.64.... \approx 15°\) | A1 | From correct working. The question asks for the answer to the nearest degree. |
## Question 4:
**Horizontal axis moments:**
$6m \times 3a + 4m \times 4a + 5m \times 2a = 15m \times y$ | M1 | Moments about horizontal axis. Terms dimensionally consistent. Condone slip with $a$. Needs all terms. Condone sign errors.
$(44ma = 15my)$ | A1 | Correct unsimplified
$y = \frac{44a}{15}$ from $B$ | A1 | Or equivalent. Correct for their axis: $\frac{46a}{15}$ from $A$, $\frac{16a}{15}$ from $E$ (CD)
**Vertical axis moments:**
$5m \times \frac{3a}{2} + 4m \times a = 15mx$ | M1 | Moments about vertical axis. Terms dimensionally consistent. Condone slip with $a$. Needs all terms. Condone sign errors.
$\left(\frac{23ma}{2} = 15mx\right)$ | A1 | Correct unsimplified
$x = \frac{23a}{30}$ from $E$ ($AB$) | A1 | Or equivalent. Correct for their axis: $\frac{67a}{30}$ from $C$
**Angle:**
$\tan\theta = \dfrac{\frac{23}{30}}{\frac{44}{15}}$ | M1 | Find a relevant angle using distances measured from $B$. Allow for $\tan\theta = \frac{88}{23}$
$= \frac{23}{88}(= 0.261...)$ | A1ft | Correct for their distances from $B$. $\binom{\text{horizontal}}{\text{vertical}}$
$\theta = 14.64.... \approx 15°$ | A1 | From correct working. The question asks for the answer to the nearest degree.
**[9]**
---4.
\begin{figure}[h]
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\includegraphics[alt={},max width=\textwidth]{3eb71ecb-fa88-4cca-a2b6-bcf11f1d689b-10_517_371_260_790}
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\caption{Figure 2}
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\end{figure}
The number "4", shown in Figure 2, is a rigid framework made from three uniform rods, $A B , B C$ and $C D$, where
$$A B = 6 a , B C = 5 a \text { and } C D = 4 a$$
The point $E$ is on $A B$ and $C D$, where $B E = 4 a , C E = 3 a$ and angle $C E B = 90 ^ { \circ }$
The three rods are all made from the same material and they all lie in the same plane.
The framework is suspended from $B$ and hangs in equilibrium with $B A$ at an angle $\theta$ to the downward vertical.
Find $\theta$ to the nearest degree.\\
VILU SIHI NI JAIUM ION OC\\
VIUV SIHI NI JAHM ION OC\\
VIIV SIHI NI EIIIM ION OC
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VIIV SIHI NI III HM ION OC & VIUV SIHI NI JIHM I ON OO & VI4V SIHI NI JIIIM ION OO \\
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\hfill \mbox{\textit{Edexcel M2 2021 Q4 [9]}}