| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2016 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Particle on cone surface – with string attached to vertex or fixed point |
| Difficulty | Standard +0.8 This is a challenging M2 conical pendulum problem requiring resolution of forces in two directions (radial and vertical), geometric reasoning with the cone angle and string angle, and careful algebraic manipulation to derive a specific expression. Part (ii) requires understanding that contact is lost when the normal force becomes zero. The multi-step nature, 3D geometry, and need to handle both constraint forces (tension and normal) simultaneously make this significantly harder than average A-level questions, though it remains a standard M2 topic. |
| Spec | 3.03e Resolve forces: two dimensions3.03m Equilibrium: sum of resolved forces = 06.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| \(T \cos 30 + R \sin 60 = mg\) | M1* | Resolving vertically (3 terms) |
| \(T \sin 30 - R \cos 60 = m(a \sin 30)\omega^2\) | M1* | Resolving horizontally (3 terms); an \(r\) used where \(r\) is not just \(a\) |
| M1 dep* | Eliminating \(T\) and solve for \(R\) in terms of \(m\), \(g\), \(a\) and \(\omega\) | |
| \(R = \frac{1}{6}m\left(2\sqrt{3}g - 3a\omega^2\right)\) | A1 [6] | AG Correctly shown |
| Answer | Marks | Guidance |
|---|---|---|
| For using \(R = 0\) to attempt to find either \(v\) or \(T\) | M1 | Or attempt to find \(\omega\) |
| \(T = \frac{mg}{\cos 30} = 39.6\) | A1 | 39.606228... |
| \(\omega^2 = \frac{T}{ma}\), \(v = 1.19 \text{ m s}^{-1}\) | A1 [3] | 1.1893309... |
## (i)
$T \cos 30 + R \sin 60 = mg$ | M1* | Resolving vertically (3 terms)
$T \sin 30 - R \cos 60 = m(a \sin 30)\omega^2$ | M1* | Resolving horizontally (3 terms); an $r$ used where $r$ is not just $a$
| M1 dep* | Eliminating $T$ and solve for $R$ in terms of $m$, $g$, $a$ and $\omega$
$R = \frac{1}{6}m\left(2\sqrt{3}g - 3a\omega^2\right)$ | A1 [6] | AG Correctly shown
## (ii)
For using $R = 0$ to attempt to find either $v$ or $T$ | M1 | Or attempt to find $\omega$
$T = \frac{mg}{\cos 30} = 39.6$ | A1 | 39.606228...
$\omega^2 = \frac{T}{ma}$, $v = 1.19 \text{ m s}^{-1}$ | A1 [3] | 1.1893309...
---A smooth solid cone of semi-vertical angle $60°$ is fixed to the ground with its axis vertical. A particle $P$ of mass $m$ is attached to one end of a light inextensible string of length $a$. The other end of the string is attached to a fixed point vertically above the vertex of the cone. $P$ rotates in a horizontal circle on the surface of the cone with constant angular velocity $\omega$. The string is inclined to the downward vertical at an angle of $30°$ (see diagram).
\begin{enumerate}[label=(\roman*)]
\item Show that the magnitude of the contact force between the cone and the particle is $\frac{1}{4}m(2\sqrt{3}g - 3a\omega^2)$. [6]
\item Given that $a = 0.5$ m and $m = 3.5$ kg, find, in either order, the greatest speed for which the particle remains in contact with the cone and the corresponding tension in the string. [3]
\end{enumerate}
\hfill \mbox{\textit{OCR M2 2016 Q4 [9]}}