M2 June 2013 Q4
4.
The uniform lamina \(ABCDEF\) is a regular hexagon with centre \(O\) and sides of length 2 m, as shown in Figure 1.
The triangles \(OAF\) and \(OEF\) are removed to form the uniform lamina \(OABCDE\), shown in Figure 2.
The lamina \(OABCDE\) is freely suspended from \(E\) and hangs in equilibrium.
| Scheme | Marks | |||||||||
|---|---|---|---|---|---|---|---|---|---|---|
 | ||||||||||
| B1 B1 | |||||||||
| \(2\bar{y} = 1 \times \dfrac{1}{2} + 1 \times \dfrac{1}{2}\) | M1 A1 | |||||||||
| \(\bar{y} = 0.5\) (m) | A1 | |||||||||
| (5) |
Notes
For a valid division into basic elements: e.g. pair of rhombuses
B1 Correct mass ratios for parts and the arrow shape
B1 Correct vertical distances from a horizontal axis
M1 Moments equation about a horizontal axis
A1 Correct equation for their axis
a alt 2
| B1 B1 | ||||||||||||
| \(4\bar{y} = 2 \times 1\) | M1A1 | ||||||||||||
| \(\bar{y} = 0.5\) (m) | A1 |
B1 B1 Rhombus + two triangles
M1A1 Moments equation
a alt 3
| B1 B1 | |||||||||
| \(4\bar{y} = 0 - -2 \times 1\) | M1A1 | |||||||||
| \(\bar{y} = 0.5\) (m) | A1 |
B1 B1 Hexagon – rhombus
a alt 4
| \(h\) = height of each triangle \(= \sqrt{3}\) Distances of c of m from horizontal through \(O\) | ||||||||||||||||
| B1 B1 | |||||||||||||||
| \(4\bar{y} = 2 \times 1 \times \dfrac{2\sqrt{3}}{3}\cos 30\ \left(= \dfrac{4\sqrt{3}}{3} \times \dfrac{\sqrt{3}}{2} = 2\right)\) | M1A1 | |||||||||||||||
| \(\bar{y} = 0.5\) (m) | A1 |
4 triangles
In 4(b) the first two marks are
M1: Indentify a triangle, with one angle correct, and attempt to find the lengths of two sides
A1ft: 2 sides correct, follow their answer to (a)
DM1: Work sufficient to be able to go on to find the required angle. Dependent on the preceding M1
A1ft: follow their answer to (a)
DM1: Find the required angle. Dependent on the preceding M1
A1 Correct answer
…. for example ……
| Scheme | Marks |
|---|---|
 | |
| \(2\cos 30 = \sqrt{3}\), "0.5"\( + 2\sin 30 = 1.5\) | M1A1ft |
| \(\tan\theta = \dfrac{\text{their }1.5}{\text{their }\sqrt{3}}\) | DM1 A1ft |
| Required angle \(= \theta - 30 = \tan^{-1}\dfrac{1.5}{\sqrt{3}} - 30 = 40.89\ldots - 30 = 10.9^\circ\) | DM1 A1 |
| (6) | |
| (11 marks) |
Notes
M1A1ft Their 0.5 & their \(\sqrt{3}\)
DM1 Use of tan in a right angled triangle. Accept the reciprocal
A1ft Correct for their angle. Ft their 0.5
DM1 Correct strategy to find required angle e.g. "\(\theta\)"\(- 30^\circ\) or \(90^\circ - 30^\circ -\)"\(\theta\)"
A1 Accept \(11^\circ\), \(10.9^\circ\) or better
4b alt
 | |
| SAS in a relevant triangle | M1A1ft |
| \(d^2 = 2^2 + 0.5^2 - 2 \times 2 \times 0.5\cos 120 = 5.25\) | DM1 A1ft |
| \(\dfrac{\sin\theta}{0.5} = \dfrac{\sin 120}{\sqrt{5.25}}\) | DM1 |
| \(\theta = 10.9^\circ\) | A1 |
M1A1ft Their 0.5
DM1 Correct cosine rule.
A1ft Correct equation. Their 0.5