| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2018 |
| Session | Specimen |
| Marks | 8 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Particle in hemispherical bowl |
| Difficulty | Standard +0.3 This is a standard M3 circular motion problem requiring resolution of forces (normal reaction and weight) and application of F=mrω². The geometry is straightforward (given depth leads to radius √3r/2 by Pythagoras), and the 'show that' format provides a clear target. Slightly easier than average due to the structured setup and standard technique application. |
| Spec | 6.05c Horizontal circles: conical pendulum, banked tracks |
# Question 1
**M1:** Resolving vertically $30°$ or $\theta$
$R\sin30° = mg$
**A1:** Correct equation $30°$ or $\theta$
**M1:** Attempting an equation of motion along the radius, acceleration in either form $30°$ or $\theta$. Allow with $r$ for radius.
$R\cos30° = m\omega r\cos30°\omega^2$
**A1:** LHS correct $30°$ or $\theta$
**A1:** RHS correct, $30°$ or $\theta$ but not $r$ for radius.
**DM1:** Obtaining an expression for $\omega^2$ or for $v^2$ and the length of the path $30°$ or $\theta$. Dependent on both previous M marks.
$\omega^2 = \frac{g}{r\sin30°} = \frac{2g}{r}$
**A1:** Correct expression for $\omega$. Must have the numerical value for the trig function now.
$\omega = \sqrt{\frac{2g}{r}}$
**A1 cso:** Deducing the GIVEN answer.
$\text{Time} = \frac{2\pi}{\omega} = \frac{2\pi}{\sqrt{\frac{2g}{r}}} = \pi\sqrt{\frac{2r}{g}}$
**(8 marks)**
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## Alternative:
**M2:** Resolve perpendicular to the reaction
$mg\cos30° = m\omega_{\text{rad}}\omega^2\cos60°$
**A1:** (LHS)
**A1:** (RHS)
$= mr\cos30°\omega^2\cos60°$
**M1:** Obtain $\omega$
**A1:** Correct $\omega$
**A1:** Correct time
**(8 marks)**1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{bb73b211-7629-4ed7-9b71-91841c29bb85-02_397_526_561_715}
\captionsetup{labelformat=empty}
\caption{Figure 1}
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\end{figure}
A hemispherical bowl, of internal radius $r$, is fixed with its circular rim upwards and horizontal. A particle $P$ of mass $m$ moves on the smooth inner surface of the bowl. The particle moves with constant angular speed in a horizontal circle. The centre of the circle is at a distance $\frac { 1 } { 2 } r$ vertically below the centre of the bowl, as shown in Figure 1.\\
The time taken by $P$ to complete one revolution of its circular path is $T$.\\
Show that $T = \pi \sqrt { \frac { 2 r } { g } }$.\\
\begin{center}
\end{center}
\hfill \mbox{\textit{Edexcel M3 2018 Q1 [8]}}