M3 June 2006 Q2
2. A bowl consists of a uniform solid metal hemisphere, of radius \(a\) and centre \(O\), from which is removed the solid hemisphere of radius \(\tfrac{1}{2}a\) with the same centre \(O\).
(a) Show that the distance of the centre of mass of the bowl from \(O\) is \(\dfrac{45}{112}a\). (5)
The bowl is fixed with its plane face uppermost and horizontal. It is now filled with liquid. The mass of the bowl is \(M\) and the mass of the liquid is \(kM\), where \(k\) is a constant. Given that the distance of the centre of mass of the bowl and liquid together from \(O\) is \(\dfrac{17}{48}a\),
(b) find the value of \(k\). (5)
| Scheme | Marks |
|---|---|
| \(\begin{array}{lccc} & \text{Small Hemisphere} & \text{Bowl} & \text{Large Hemisphere} \\ \text{Mass ratios} & \frac{2}{3}\pi\left(\frac{a}{2}\right)^3 & \frac{2}{3}\pi\frac{7a^3}{8} & \frac{2}{3}\pi a^3 \end{array}\) Anything in the ratio \(1:7:8\) | B1 |
| \(\begin{array}{lccc} \bar{x} & \frac{3}{16}a & \bar{x} & \frac{3}{8}a \end{array}\) | B1 |
| \(1 \times \dfrac{3}{16}a + 7 \times \bar{x} = 8 \times \dfrac{3}{8}a\) | M1 A1 |
| Leading to \(\bar{x} = \dfrac{45}{112}a\) * cso | A1 |
| (5) |
| Scheme | Marks |
|---|---|
| \(\begin{array}{lccc} & \text{Bowl} & \text{Liquid} & \text{Bowl and Liquid} \\ \text{Mass Ratios} & M & kM & (k+1)M \end{array}\) | B1 |
| \(\begin{array}{lccc} \bar{x} & \frac{45}{112}a & \frac{3}{16}a & \frac{17}{48}a \end{array}\) | B1 |
| \(M \times \dfrac{45}{112}a + kM \times \dfrac{3}{16}a = (k+1)M \times \dfrac{17}{48}a\) | M1 A1 |
| Leading to \(k = \dfrac{2}{7}\) | A1 |
| (5) | |
| (10 marks) |