| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2017 |
| Session | January |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Solid with removed cylinder or hemisphere from solid |
| Difficulty | Standard +0.3 This is a standard M3 centre of mass problem using the composite body method (original solid minus removed part). It requires knowing standard results for cylinder and hemisphere centres of mass, then applying the formula with careful bookkeeping of signs and distances. The calculation is straightforward with no geometric insight needed beyond the setup. |
| Spec | 6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Ratio of masses: \(96\pi a^3\rho : 18\pi a^3\rho : 78\pi a^3\rho\), i.e. \(16 : 3 : 13\) | M1A1 | Attempt ratio; terms consistent in \(\pi, a^3, \rho\); formulae correct. Simplification not required. |
| Distance from \(O\): \(3a \quad \frac{9a}{8} \quad \bar{x}\) | B1 | Correct distances from \(O\) or any other point. |
| \(16\times 3a - 3\times\frac{9a}{8} = 13\bar{x}\) | M1A1ft | Form moments equation; dimensionally correct; hemisphere subtracted on LHS. Correct equation following their values. |
| \(\bar{x} = \frac{357}{104}a = 3\frac{45}{104}a\) or \(3.432a...\) (accept \(3.4a\) or better) | A1 [6] | Correct answer, exact or decimal; must be from \(O\) and positive. |
# Question 3:
| Working/Answer | Marks | Guidance |
|---|---|---|
| Ratio of masses: $96\pi a^3\rho : 18\pi a^3\rho : 78\pi a^3\rho$, i.e. $16 : 3 : 13$ | M1A1 | Attempt ratio; terms consistent in $\pi, a^3, \rho$; formulae correct. Simplification not required. |
| Distance from $O$: $3a \quad \frac{9a}{8} \quad \bar{x}$ | B1 | Correct distances from $O$ or any other point. |
| $16\times 3a - 3\times\frac{9a}{8} = 13\bar{x}$ | M1A1ft | Form moments equation; dimensionally correct; hemisphere subtracted on LHS. Correct equation following their values. |
| $\bar{x} = \frac{357}{104}a = 3\frac{45}{104}a$ or $3.432a...$ (accept $3.4a$ or better) | A1 [6] | Correct answer, exact or decimal; must be from $O$ and positive. |
---3.
\begin{figure}[h]
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\includegraphics[alt={},max width=\textwidth]{85d8fc7d-8863-419e-8eef-8751a6fb6315-04_647_684_260_635}
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\caption{Figure 2}
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A uniform right circular solid cylinder has radius $4 a$ and height $6 a$. A solid hemisphere of radius $3 a$ is removed from the cylinder forming a solid $S$. The upper plane face of the cylinder coincides with the plane face of the hemisphere. The centre of the upper plane face of the cylinder is $O$ and this is also the centre of the plane face of the hemisphere, as shown in Figure 2. Find the distance from $O$ to the centre of mass of $S$.\\
(6)\\
\hfill \mbox{\textit{Edexcel M3 2017 Q3 [6]}}