M2 June 2015 Q2
2.
The uniform lamina \(OABCD\), shown in Figure 1, is formed by removing the triangle \(OAD\) from the square \(ABCD\) with centre \(O\). The square has sides of length \(2a\).
The mass of the lamina is \(M\). A particle of mass \(kM\) is attached to the lamina at \(D\) to form the system \(S\). The system \(S\) is freely suspended from \(A\) and hangs in equilibrium with \(AO\) vertical.
| Scheme | Marks |
|---|---|
| Ratio of masses 1 : 3 : 4 | B1 |
| Centre of mass of triangle \(\dfrac{2}{3}a\) from \(O\) | B1 |
| Moments about horizontal axis through \(O\): \(\ 4 \times 0 - 1 \times \dfrac{2}{3}a = 3d\) | M1 |
| \(\left(d = -\dfrac{2}{9}a\right)\ \ \) Distance \(= \dfrac{2}{9}a\) | A1 |
| (4) |
Notes
B1 Correct ratios for their division. Also common: 3 equal triangles, 6 equal triangles, a rectangle and two equal triangles
B1 Correct centres for the triangles in their division consistent with their axis.
M1 Condone \(4 \times 0\) not seen. Terms must be of correct form. Condone use of moments about a parallel axis. Signs must be consistent with their axis and distances. Watch out for people who add a triangle to the square.
A1 Reach *Given answer* with no errors seen. Their answer must be positive
| Scheme | Marks |
|---|---|
| \(\sqrt{2}akM = \dfrac{2}{9}a\cos 45M\) | M1 A2 |
| \(k = \dfrac{1}{9}\) | A1 |
| (4) | |
| (8 marks) |
Notes
M1 Moments about \(A\). Lengths must be resolved as necessary (need use of trig).
A2 -1 each error
Alt 2b
| Take \(A\) as origin and axes along \(AD\) and \(AB\), use moments to find components of distance of c of m from \(A\). | M1 |
| \(\bar{x} = \dfrac{a(1 + 2k)}{1 + k}\) | A1 |
| \(\bar{y} = \dfrac{11a}{9(1 + k)}\) | A1 |
| \(\bar{x} = \bar{y} \Rightarrow \dfrac{11}{9} = 1 + 2k \Rightarrow k = \dfrac{1}{9}\) | A1 |
M1 Could choose a different origin
A1 CWO. This mark is not available if they have assumed that the c of m of the system is at \(O\).
Alt 2b
| \(\dfrac{2}{9}a\cos 45 = \dfrac{1}{9}.\sqrt{2}a = \dfrac{1}{9}OD\) | M1A1 |
| \(kM \times OD = M \times \dfrac{1}{9}OD\) | A1 |
| \(\Rightarrow k = \dfrac{1}{9}\) | A1 |
M1A1 Using ratios