| Exam Board | Edexcel |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2022 |
| Session | October |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Rod on peg or cylinder |
| Difficulty | Standard +0.3 This is a standard M2 moments question with a rod resting on a hemisphere. Parts (a) and (b) are 'show that' questions requiring Pythagoras and taking moments about A respectively - routine techniques. Part (c) requires resolving forces horizontally and vertically then finding the resultant, which is straightforward application of standard methods. The geometry is given clearly and no novel insight is required, making this slightly easier than average. |
| Spec | 3.03m Equilibrium: sum of resolved forces = 03.04b Equilibrium: zero resultant moment and force |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(AD = \sqrt{(2a)^2+(5a)^2} = \sqrt{29}a\) | B1* | Correct use of Pythagoras to show given answer from correct working (need \(a\) on both sides) |
| Total: 1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(M(A)\): \(W \times 4a\cos\theta = N \times 5a\) | M1 | Dimensionally correct equation in \(a\). Allow if \(a\) cancelled. Condone sin/cos confusion |
| \(W \times 4a \times \dfrac{5}{\sqrt{29}} = N \times 5a\) | A1 | Correct unsimplified equation. Allow with \(\cos\theta\). NB: \(5a = \sqrt{29}a\cos\theta\) |
| \(N = \dfrac{4}{\sqrt{29}}W\) | A1* | Obtain given answer from correct working |
| Total: 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Resolve vertically | M1 | Requires all relevant terms. Condone sin/cos confusion |
| \(V + N\cos\theta = W\ \left(V = \dfrac{9}{29}W\right)\) or \(kW\sin\alpha + N\cos\theta = W\) | A1 | Correct unsimplified equation. Need not substitute for trig |
| Resolve horizontally | M1 | Requires all relevant terms. Condone consistent sin/cos confusion |
| \(H = N\sin\theta\left(=\dfrac{8}{29}W\right)\) or \(kW\cos\alpha = N\sin\theta\left(=\dfrac{8}{29}W\right)\) | A1 | Correct unsimplified equation. Need not substitute for trig |
| Use Pythagoras to obtain \(k\): \(k^2 = \left(\dfrac{9}{29}\right)^2 + \left(\dfrac{8}{29}\right)^2\) | DM1 | Correct use of perpendicular components. Dependent on the first 2 M marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Resolve parallel to rod | M1 | Requires all relevant terms. Condone sin/cos confusion |
| \(F = W\sin\theta \left(= \frac{2}{\sqrt{29}}W\right)\) | A1 | Correct unsimplified equation. Need not substitute for trig |
| Resolve perpendicular to rod | M1 | Requires all relevant terms. Condone consistent sin/cos confusion |
| \(E + N = W\cos\theta \left(E = \frac{1}{\sqrt{29}}W\right)\) | A1 | Correct unsimplified equation. Need not substitute for trig |
| Alt: \(aW\cos\theta + 5aH\sin\theta = 5aV\cos\theta\) or \(aW\cos\theta = kW \times 5a\sin(\alpha-\theta)\) | — | Possible alternative M(C) equations |
| Use Pythagoras to obtain \(k\) | M1 | Correct use of Pythagoras |
| \(k = \frac{1}{\sqrt{29}}\sqrt{1+4} = \sqrt{\frac{5}{29}}\) | A1 | Correct only |
| Use trig to obtain \(\tan\alpha\): \(\tan(\alpha-\theta) = \frac{1}{2} = \frac{\tan\alpha - \frac{2}{5}}{1 + \frac{2}{5}\tan\alpha}\) | DM1 | Use of trig to obtain expression in \(\tan\alpha\) |
| \(\tan\alpha = \frac{9}{8}\) | A1 | Correct only |
| Total | 8 |
# Question 5:
## Part (a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $AD = \sqrt{(2a)^2+(5a)^2} = \sqrt{29}a$ | B1* | Correct use of Pythagoras to show given answer from correct working (need $a$ on both sides) |
| **Total: 1** | | |
## Part (b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $M(A)$: $W \times 4a\cos\theta = N \times 5a$ | M1 | Dimensionally correct equation in $a$. Allow if $a$ cancelled. Condone sin/cos confusion |
| $W \times 4a \times \dfrac{5}{\sqrt{29}} = N \times 5a$ | A1 | Correct unsimplified equation. Allow with $\cos\theta$. NB: $5a = \sqrt{29}a\cos\theta$ |
| $N = \dfrac{4}{\sqrt{29}}W$ | A1* | Obtain given answer from correct working |
| **Total: 3** | | |
## Part (c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolve vertically | M1 | Requires all relevant terms. Condone sin/cos confusion |
| $V + N\cos\theta = W\ \left(V = \dfrac{9}{29}W\right)$ or $kW\sin\alpha + N\cos\theta = W$ | A1 | Correct unsimplified equation. Need not substitute for trig |
| Resolve horizontally | M1 | Requires all relevant terms. Condone consistent sin/cos confusion |
| $H = N\sin\theta\left(=\dfrac{8}{29}W\right)$ or $kW\cos\alpha = N\sin\theta\left(=\dfrac{8}{29}W\right)$ | A1 | Correct unsimplified equation. Need not substitute for trig |
| Use Pythagoras to obtain $k$: $k^2 = \left(\dfrac{9}{29}\right)^2 + \left(\dfrac{8}{29}\right)^2$ | DM1 | Correct use of perpendicular components. Dependent on the first 2 M marks |
# Question 5(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolve parallel to rod | M1 | Requires all relevant terms. Condone sin/cos confusion |
| $F = W\sin\theta \left(= \frac{2}{\sqrt{29}}W\right)$ | A1 | Correct unsimplified equation. Need not substitute for trig |
| Resolve perpendicular to rod | M1 | Requires all relevant terms. Condone consistent sin/cos confusion |
| $E + N = W\cos\theta \left(E = \frac{1}{\sqrt{29}}W\right)$ | A1 | Correct unsimplified equation. Need not substitute for trig |
| Alt: $aW\cos\theta + 5aH\sin\theta = 5aV\cos\theta$ or $aW\cos\theta = kW \times 5a\sin(\alpha-\theta)$ | — | Possible alternative M(C) equations |
| Use Pythagoras to obtain $k$ | M1 | Correct use of Pythagoras |
| $k = \frac{1}{\sqrt{29}}\sqrt{1+4} = \sqrt{\frac{5}{29}}$ | A1 | Correct only |
| Use trig to obtain $\tan\alpha$: $\tan(\alpha-\theta) = \frac{1}{2} = \frac{\tan\alpha - \frac{2}{5}}{1 + \frac{2}{5}\tan\alpha}$ | DM1 | Use of trig to obtain expression in $\tan\alpha$ |
| $\tan\alpha = \frac{9}{8}$ | A1 | Correct only |
| **Total** | **8** | |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{1732eb73-8c16-4a45-8d3b-a88e659e47ea-12_424_1118_221_420}
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\caption{Figure 1}
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\end{figure}
A uniform rod $A B$ has length $8 a$ and weight $W$.\\
The end $A$ of the rod is freely hinged to horizontal ground.\\
The rod rests in equilibrium against a block which is also fixed to the ground.\\
The block is modelled as a smooth solid hemisphere with radius $2 a$ and centre $D$.\\
The point of contact between the rod and the block is $C$, where $A C = 5 a$\\
The rod is at an angle $\theta$ to the ground, as shown in Figure 1.\\
Points $A , B , C$ and $D$ all lie in the same vertical plane.
\begin{enumerate}[label=(\alph*)]
\item Show that $A D = \sqrt { 29 } a$
\item Show that the magnitude of the normal reaction at $C$ between the rod and the block is $\frac { 4 } { \sqrt { 29 } } W$
The resultant force acting on the rod at $A$ has magnitude $k W$ and acts at an angle $\alpha$ to the ground.
\item Find (i) the exact value of $k$\\
(ii) the exact value of $\tan \alpha$
\begin{center}
\end{center}
\begin{center}
\end{center}
\includegraphics[max width=\textwidth, alt={}, center]{1732eb73-8c16-4a45-8d3b-a88e659e47ea-15_72_1819_2709_114}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M2 2022 Q5 [12]}}