| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2013 |
| Session | June |
| Marks | 7 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Particle on cone surface – no string (normal reaction only) |
| Difficulty | Standard +0.3 This is a standard M3 circular motion problem requiring resolution of forces (normal reaction and weight) on a conical surface, finding the angle from geometry, then applying F=mrω². The 'show that' format and straightforward geometry make it slightly easier than average, though it requires careful coordinate resolution and algebraic manipulation across multiple steps. |
| Spec | 6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Marks | Guidance |
| Vertical: \(R\cos\beta = mg\) | M1A1 | Resolving vertically |
| Horizontal: \(R\sin\beta = \frac{mv^2}{r} = \frac{3mv^2}{a}\) | M1A1 | Resolving horizontally, using \(r = \frac{a}{3}\) |
| Divide: \(\tan\beta = \frac{3mv^2}{amg}\) | M1dep | Dependent on both previous M marks |
| \(\tan\beta = \frac{h}{a}\) | B1 | Geometric relationship for \(\tan\beta\) |
| \(\frac{3mv^2}{amg} = \frac{h}{a}\), \(\quad \frac{3v^2}{g} = h\), \(\quad v = \sqrt{\dfrac{hg}{3}}\) | A1 | *AG* — given result, must be shown completely |
| Total | (7) |
## Question 1:
| Working/Answer | Marks | Guidance |
|---|---|---|
| Vertical: $R\cos\beta = mg$ | M1A1 | Resolving vertically |
| Horizontal: $R\sin\beta = \frac{mv^2}{r} = \frac{3mv^2}{a}$ | M1A1 | Resolving horizontally, using $r = \frac{a}{3}$ |
| Divide: $\tan\beta = \frac{3mv^2}{amg}$ | M1dep | Dependent on both previous M marks |
| $\tan\beta = \frac{h}{a}$ | B1 | Geometric relationship for $\tan\beta$ |
| $\frac{3mv^2}{amg} = \frac{h}{a}$, $\quad \frac{3v^2}{g} = h$, $\quad v = \sqrt{\dfrac{hg}{3}}$ | A1 | *AG* — given result, must be shown completely |
| **Total** | **(7)** | |1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{daa795f0-2c5e-4617-a295-fbe74c22be4a-02_679_568_210_680}
\captionsetup{labelformat=empty}
\caption{Figure 1}
\end{center}
\end{figure}
A hollow right circular cone, of base radius $a$ and height $h$, is fixed with its axis vertical and vertex downwards, as shown in Figure 1. A particle moves with constant speed $v$ in a horizontal circle of radius $\frac { 1 } { 3 } a$ on the smooth inner surface of the cone.
Show that $v = \sqrt { } \left( \frac { 1 } { 3 } h g \right)$.\\
\hfill \mbox{\textit{Edexcel M3 2013 Q1 [7]}}