AS June 2022 Q2
2.
Uniform wire is used to form the framework shown in Figure 2.
In the framework
- \(ABCD\) is a rectangle with \(AD = 2a\) and \(DC = a\)
- \(BEC\) is a semicircular arc of radius \(a\) and centre \(O\), where \(O\) lies on \(BC\)
The diameter of the semicircle is \(BC\) and the point \(E\) is such that \(OE\) is perpendicular to \(BC\).
The points \(A\), \(B\), \(C\), \(D\) and \(E\) all lie in the same plane.
The framework is freely suspended from \(A\) and hangs in equilibrium with \(AE\) at an angle \(\theta^\circ\) to the downward vertical.
The mass of the framework is \(M\).
A particle of mass \(kM\) is attached to the framework at \(B\).
The centre of mass of the loaded framework lies on \(OA\).
| Scheme | Marks | AO |
|---|---|---|
| \(ABCD\) \(BEC\) framework | ||
| \(6a\) \(\pi a\) \(6a + \pi a\) | B1 | 1.2 |
| \(\dfrac{1}{2}a\) \((-)\dfrac{2a}{\pi}\) \(\bar{x}\) | B1 | 1.2 |
| Moments about \(BC\) | M1 | 2.1 |
| \(6a \times \dfrac{1}{2}a - \pi a \times \dfrac{2a}{\pi} = (6a + \pi a)\bar{x}\) | A1 | 1.1b |
| \(\bar{x} = \dfrac{a}{6+\pi}\) * | A1* | 2.2a |
| (5) |
Notes
B1: Any equivalent ratios
B1: Or correct distances from a parallel axis
M1: Or moments about a parallel axis
Must be using framework. If \(BC\) included twice mark as a misread.
A1: Correct unsimplified equation for their axis.
Allow within a vector equation
A1*: Correct given answer correctly obtained
| Scheme | Marks | AO |
|---|---|---|
| Angle \(DAE = \tan^{-1}\left(\dfrac{2a}{a}\right)\) | M1 | 1.1b |
| Angle \(DAG = \tan^{-1}\left(\dfrac{a - \dfrac{a}{6+\pi}}{a}\right) = \tan^{-1}\left(\dfrac{5+\pi}{6+\pi}\right)\) | M1 | 1.1b |
| Angle \(= DAE - DAG\) | M1 | 3.1a |
| \(21.74637\ldots\) | A1 | 1.1b |
| (4) |
Notes
M1: Correct relevant angle (or side if they use the cosine rule) Do not need to evaluate: accept \(\tan\alpha = \ldots\) or \(\alpha = \tan^{-1}\ldots\) (e.g. \(63.4\ldots^\circ\) or \(90^\circ - 63.4\ldots^\circ\))
M1: Another correct relevant angle (or side if they use the cosine rule) Do not need to evaluate: accept \(\tan\beta = \ldots\) or \(\beta = \tan^{-1}\ldots\) (e.g. \(41.68\ldots^\circ\) or \(90^\circ - 41.68\ldots^\circ\))
M1: Correct method for finding the required angle
A1: \(22^\circ\) or better
| Scheme | Marks | AO |
|---|---|---|
| Moments about \(OA\) | M1 | 2.1 |
| \(kMa\sin 45^\circ = M\bar{x}\sin 45^\circ\) | A1 | 1.1b |
| \(k = \dfrac{1}{6+\pi}\) \((= 0.10939\ldots)\) | A1 | 1.1b |
| (3) | ||
| (12 marks) |
Notes
M1: Complete method to give an equation in \(k\) only
A1: Correct equation in \(k\) only
A1: 0.11 or better
Alternative (c)
| Scheme | Marks | AO |
|---|---|---|
| Moments about \(O\) | M1 | 2.1 |
| \(kM\begin{pmatrix} 0 \\ a \end{pmatrix} - M\begin{pmatrix} \frac{a}{6+\pi} \\ 0 \end{pmatrix} = (k+1)M\begin{pmatrix} -\lambda \\ \lambda \end{pmatrix}\) | A1 | 1.1b |
| \(k = \dfrac{1}{6+\pi}\) \((= 0.10939\ldots)\) | A1 | 1.1b |
| (3) |