AS June 2024 Q1
1.
A uniform rod of length \(24a\) is cut into seven pieces which are used to form the framework \(ABCDEF\) shown in Figure 1.
It is given that
- \(AF = BE = CD = AB = FE = 4a\)
- \(BC = ED = 2a\)
- the rods \(AF\), \(BE\) and \(CD\) are parallel
- the rods \(AB\), \(BC\), \(FE\) and \(ED\) are parallel
- \(AF\) is perpendicular to \(AB\)
- the rods all lie in the same plane
The distance of the centre of mass of the framework from \(AF\) is \(d\).
(a) Show that \(d = \dfrac{19}{6}a\) (4)
(b) Find the distance of the centre of mass of the framework from \(A\). (3)
| Scheme | Marks | AO |
|---|---|---|
| M(\(AF\)) | M1 | 2.1 |
| \(24ad = 4a \times 4a + 4a \times 6a + 2 \times 3a \times 6a\ \left(= 76a^2\right)\) | A1 A1 | 1.1b 1.1b |
| \(d = \dfrac{19a}{6}\) * | A1* | 2.2a |
| (4) |
Notes
M1: Moments about \(AF\) or a parallel axis. All terms required. Dimensionally correct but allow consistent cancelling of a factor of \(a\) or \(2a\).
A1: Unsimplified equation with at most one error
A1: Correct unsimplified equation
E.g. \(12d = 2 \times 4a + 2 \times 6a + 2 \times 3 \times 3a\ (= 38a)\)
A1*: Obtain given answer including “\(d =\)” from correct exact working
| Scheme | Marks | AO |
|---|---|---|
| \(\bar{y} = 2a\) | B1 | 1.1b |
| \(D^2 = \text{their } \bar{x}^2 + \text{their } \bar{y}^2\ \left(= \dfrac{361}{36}a^2 + 4a^2\right)\) | M1 | 1.1b |
| \(D = \sqrt{\dfrac{505}{36}}a = \dfrac{\sqrt{505}}{6}a\) | A1 | 1.1b |
| (3) | ||
| (7 marks) |
Notes
B1: Seen or implied
M1: Correct use of Pythagoras to find \(D\) or \(D^2\)
A1: Any equivalent form. Accept \(3.7a\) or better. (3.7453675…)