AS June 2025 Q1
1.
A uniform plane lamina, shown shaded in Figure 1, is formed by removing an isosceles triangle \(ABE\) from a rectangle \(ABCD\).
- The midpoint of \(AB\) is \(M\)
- \(AB = 3a\)
- \(AD = 4a\)
- \(EM = 3a\)
- \(AE = EB\)
(a) Show that the distance of the centre of mass of the lamina from \(AB\) is \(\dfrac{13a}{5}\) (5)
The lamina is suspended by a string attached to the lamina at \(D\).
The lamina hangs freely in equilibrium.
(b) Find, to the nearest degree, the angle between \(AD\) and the vertical. (3)
| Scheme | Marks | AO |
|---|---|---|
| \(ABCD\) \(ABE\) lamina | ||
| \(12a^2\) \(\dfrac{9a^2}{2}\) \(\dfrac{15a^2}{2}\) | B1 | 1.2 |
| \(2a\) \(a\) \(\bar{x}\) | B1 | 1.2 |
| Moments about \(AB\) | M1 | 2.1 |
| \((12a^2 \times 2a) - \left(\dfrac{9a^2}{2} \times a\right) = \dfrac{15a^2}{2}\bar{x}\) | A1 | 1.1b |
| \(\bar{x} = \dfrac{13a}{5}\) * | A1* | 2.2a |
| (5) |
Notes
B1: Any equivalent ratios
B1: Or correct distances from a parallel axis
M1: Or moments about a parallel axis
A1: Correct unsimplified equation for their axis
A1*: Correct given answer correctly obtained. Condone \(\dfrac{13}{5}a\)
| Scheme | Marks | AO |
|---|---|---|
| Use of trigonometry to find a relevant angle | M1 | 3.1a |
| \(\tan\alpha = \dfrac{\dfrac{3a}{2}}{4a - \dfrac{13a}{5}}\) | A1 | 1.1b |
| \(\alpha = 47^\circ\) or \(133^\circ\) (nearest degree) | A1 | 1.1b |
| (3) | ||
| (8 marks) |
Notes
M1: A complete method to obtain angle with e.g. vertical or horizontal
A1: Correct unsimplified equation (or its reciprocal)
A1: cao