A2 October 2021 Q3
3.

A uniform solid hemisphere \(H\) has radius \(2a\). A solid hemisphere of radius \(a\) is removed from the hemisphere \(H\) to form a bowl. The plane faces of the hemispheres coincide and the centres of the two hemispheres coincide at the point \(O\), as shown in Figure 2.
The centre of mass of the bowl is at the point \(G\).
Figure 3 below shows a cross-section of the bowl which is resting in equilibrium with a point \(P\) on its curved surface in contact with a rough plane. The plane is inclined to the horizontal at an angle \(\alpha\) and is sufficiently rough to prevent the bowl from slipping. The line \(OG\) is horizontal and the points \(O\), \(G\) and \(P\) lie in a vertical plane which passes through a line of greatest slope of the inclined plane.

| Scheme | Marks | AO |
|---|---|---|
| Mass ratios: \(\dfrac{2}{3}\pi(2a)^3,\ \dfrac{2}{3}\pi a^3,\ \dfrac{2}{3}\pi(2a)^3 - \dfrac{2}{3}\pi a^3 \qquad (8, 1, 7)\) | B1 | 1.2 |
| Distances: \(\dfrac{3}{8}(2a),\ \dfrac{3}{8}a,\ \bar{x}\) | B1 | 1.2 |
| \(\left(\dfrac{2}{3}\pi(2a)^3 - \dfrac{2}{3}\pi a^3\right)\bar{x} = \dfrac{2}{3}\pi(2a)^3\times\dfrac{3}{8}(2a) - \dfrac{2}{3}\pi a^3\times\dfrac{3}{8}a\) | M1 | 3.1a |
| \(\bar{x} = \dfrac{45a}{56}\) * | A1* | 2.2a |
| (4) |
Notes
B1: Correct unsimplified (8, 1, 7)
B1: Correct unsimplified but distances could be measured from a parallel axis
M1: All terms needed and must be dimensionally correct
A1*: Correct answer correctly derived
| Scheme | Marks | AO |
|---|---|---|
| Use of appropriate trig ratio e.g. \(\sin\alpha = \dfrac{\frac{45a}{56}}{2a}\) | M1 | 3.1a |
| \(\alpha = 23.7^\circ\) (3 sf) | A1 | 1.1b |
| (2) | ||
| (6 marks) |
Notes
M1: Must be using an appropriate trig ratio
A1: Cao to 3SF