| Exam Board | CAIE |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2003 |
| Session | November |
| Marks | 6 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Moments |
| Type | Toppling and sliding of solids |
| Difficulty | Standard +0.3 This is a standard toppling problem requiring identification of the critical angle when the vertical line through the center of mass passes through the pivot point. The geometry is straightforward (cone center of mass at h/4), and part (ii) requires understanding that μ ≥ tan θ for toppling before sliding. While it involves multiple concepts (moments, friction, geometry), it follows a well-established method with no novel insight required. |
| Spec | 6.04e Rigid body equilibrium: coplanar forces |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| For stating or implying that the centre of mass is vertically above the lowest point of the cone, and with \(\bar{y} = 5\) | B1 | |
| For using \(\tan\theta = \frac{10}{y}\) or equivalent | M1 | |
| \(\theta = 63.4°\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| For using \(F < \mu R\) | M1 | |
| \(mg\sin\theta < \mu mg\cos\theta\) | A1 | |
| For using \(\mu = \tan\phi\) where \(\phi\) is the angle of friction; \(\phi > \theta\) because cone topples without sliding | M1, A1 | Alternative method |
| Coefficient is greater than 2 (ft on \(\tan\theta\) in (i)) | A1ft | N.B. Direct quotation of "topples if \(\mu > \tan\theta\)" scores B2; \(\mu > 2\) scores B1 |
# Question 2(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| For stating or implying that the centre of mass is vertically above the lowest point of the cone, and with $\bar{y} = 5$ | B1 | |
| For using $\tan\theta = \frac{10}{y}$ or equivalent | M1 | |
| $\theta = 63.4°$ | A1 | |
# Question 2(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| For using $F < \mu R$ | M1 | |
| $mg\sin\theta < \mu mg\cos\theta$ | A1 | |
| For using $\mu = \tan\phi$ where $\phi$ is the angle of friction; $\phi > \theta$ because cone topples without sliding | M1, A1 | Alternative method |
| Coefficient is greater than 2 (ft on $\tan\theta$ in (i)) | A1ft | N.B. Direct quotation of "topples if $\mu > \tan\theta$" scores B2; $\mu > 2$ scores B1 |
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\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{be83d46f-bf5b-4382-b424-bb5067626adc-2_376_569_559_466}
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\caption{Fig. 1}
\end{center}
\end{figure}
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{be83d46f-bf5b-4382-b424-bb5067626adc-2_485_456_450_1226}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
A uniform solid cone has height 20 cm and base radius 10 cm . It is placed with its axis vertical on a rough horizontal plane (see Fig. 1). The plane is slowly tilted and the cone remains in equilibrium until the angle of inclination of the plane reaches $\theta ^ { \circ }$, when the cone begins to topple without sliding (see Fig. 2).\\
(i) Find the value of $\theta$.\\
(ii) What can you say about the value of the coefficient of friction between the cone and the plane?
\hfill \mbox{\textit{CAIE M2 2003 Q2 [6]}}