| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2013 |
| Session | January |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Centre of Mass 1 |
| Type | Composite solid with hemisphere and cylinder/cone |
| Difficulty | Standard +0.8 This is a substantial M3 centre of mass problem requiring: (a) deriving a formula for composite solid COM using standard results for cone and hemisphere, involving algebraic manipulation with parameter k; (b) applying equilibrium conditions with the solid suspended at angle, requiring geometric reasoning to relate the angle to k. The multi-step nature, algebraic complexity, and need to connect geometry with mechanics principles makes this harder than average, though it uses standard M3 techniques throughout. |
| Spec | 6.04c Composite bodies: centre of mass6.04d Integration: for centre of mass of laminas/solids6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| Mass: \(\frac{2}{3}\pi r^3\) and \(\frac{1}{3}k\pi r^3\), totalling \(2+k\) | B1 | |
| Dist from \(O\): \(-\frac{3}{8}r\) and \(\frac{1}{4}kr\), combined \(\bar{x}\) | B1 | |
| \(-\frac{3}{4}r + \frac{k^2r}{4} = \bar{x}(2+k)\) | M1A1ft | |
| \(\bar{x} = \frac{(k^2-3)r}{4(k+2)}\) | A1 | Given/shown result |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(\tan\theta = \frac{(k^2-3)r}{4(k+2)} \div r\) | M1A1 | |
| \(\frac{(k^2-3)}{4(k+2)} = \frac{11}{14}\) | ||
| \(14k^2 - 42 = 44k + 88\) | ||
| \(7k^2 - 22k - 65 = 0\) | ||
| \((7k+13)(k-5) = 0\) | ||
| \(k = 5\) | M1depA1 |
## Question 2:
### Part (a):
| Answer/Working | Marks | Notes |
|---|---|---|
| Mass: $\frac{2}{3}\pi r^3$ and $\frac{1}{3}k\pi r^3$, totalling $2+k$ | B1 | |
| Dist from $O$: $-\frac{3}{8}r$ and $\frac{1}{4}kr$, combined $\bar{x}$ | B1 | |
| $-\frac{3}{4}r + \frac{k^2r}{4} = \bar{x}(2+k)$ | M1A1ft | |
| $\bar{x} = \frac{(k^2-3)r}{4(k+2)}$ | A1 | Given/shown result |
### Part (b):
| Answer/Working | Marks | Notes |
|---|---|---|
| $\tan\theta = \frac{(k^2-3)r}{4(k+2)} \div r$ | M1A1 | |
| $\frac{(k^2-3)}{4(k+2)} = \frac{11}{14}$ | | |
| $14k^2 - 42 = 44k + 88$ | | |
| $7k^2 - 22k - 65 = 0$ | | |
| $(7k+13)(k-5) = 0$ | | |
| $k = 5$ | M1depA1 | |
---2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d19c7390-0332-4cab-82e5-72976bd499a2-03_636_529_322_662}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A uniform solid consists of a right circular cone of radius $r$ and height $k r$, where $k > \sqrt { } 3$, fixed to a hemisphere of radius $r$. The centre of the plane face of the hemisphere is $O$ and this plane face coincides with the base of the cone, as shown in Figure 1.
\begin{enumerate}[label=(\alph*)]
\item Show that the distance of the centre of mass of the solid from $O$ is
$$\frac { \left( k ^ { 2 } - 3 \right) r } { 4 ( k + 2 ) }$$
The point $A$ lies on the circumference of the base of the cone. The solid is suspended by a string attached at $A$ and hangs freely in equilibrium. The angle between $A O$ and the vertical is $\theta$, where $\tan \theta = \frac { 11 } { 14 }$
\item Find the value of $k$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2013 Q2 [9]}}