| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2011 |
| Session | January |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Conical pendulum (horizontal circle) |
| Difficulty | Standard +0.3 This is a standard M3 circular motion question with two familiar scenarios: conical pendulum and particle on sphere. Part (a) requires routine resolution of forces and circular motion formula (T cos θ = mg, T sin θ = mv²/r). Part (b) uses energy conservation and Newton's second law in standard ways. The 'show that' in (b)(i) is straightforward application of energy conservation. While multi-part with several steps, all techniques are textbook exercises requiring no novel insight—slightly easier than average A-level. |
| Spec | 6.05c Horizontal circles: conical pendulum, banked tracks6.05d Variable speed circles: energy methods6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(T\cos\alpha = m\frac{V^2}{r}\) (\(\alpha\) is angle APC) | M1 | Equation of motion including \(\frac{V^2}{r}\) |
| \(T\times\frac{8.4}{30} = 48\times\frac{3.5^2}{8.4}\) | A1 | Or \(T\cos73.7=\ldots\) or \(T\sin16.3=\ldots\) |
| Tension is 250 N | A1 | |
| \(T\sin\alpha + R = mg\) | M1 | Resolving vertically (three terms) |
| \(250\times0.96 + R = 48\times9.8\) | ||
| Normal reaction is 230.4 N | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(T\sin\alpha = mg\) | M1 | Vertical equation with \(R=0\) |
| \(T\times0.96 = 48\times9.8\) | A1 | Or \(T\sin73.7=\ldots\) or \(T\cos16.3=\ldots\) |
| \(T = 490\) | ||
| \(490\times0.28 = 48\times\frac{V^2}{8.4}\) | M1 | Obtaining equation for \(V\) |
| \(V = 4.9\) | A1 | Allow \(T=490\) obtained in (i) and used correctly in (ii) for full marks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(\frac{1}{2}m(v^2-u^2) = m\times9.8(2.5-2.5\cos\theta)\) | M1 | Equation involving KE and PE |
| \(v^2 - u^2 = 49(1-\cos\theta)\) | A1 | |
| \(v^2 = u^2 + 49 - 49\cos\theta\) | E1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(mg\cos\theta - R = m\frac{v^2}{r}\) | M1, A1 | Radial equation (three terms) |
| \(48\times9.8\left(\frac{u^2+49-v^2}{49}\right) - R = \frac{48v^2}{2.5}\) | M1 | Obtaining equation in \(R\), \(u\), \(v\) |
| \(9.6u^2 + 470.4 - 9.6v^2 - R = 19.2v^2\) | ||
| \(R = 470.4 + 9.6u^2 - 28.8v^2\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(470.4 + 9.6u^2 - 28.8\times4.15^2 = 0\) | M1 | Substituting \(R=0\) and \(v=4.15\) or other complete method leading to equation for \(u\) |
| \(u = 1.63\) (3 sf) | A1 | ft requires \(0 < u < 4.15\) |
# Question 2:
## Part (a)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $T\cos\alpha = m\frac{V^2}{r}$ ($\alpha$ is angle APC) | M1 | Equation of motion including $\frac{V^2}{r}$ |
| $T\times\frac{8.4}{30} = 48\times\frac{3.5^2}{8.4}$ | A1 | Or $T\cos73.7=\ldots$ or $T\sin16.3=\ldots$ |
| Tension is 250 N | A1 | |
| $T\sin\alpha + R = mg$ | M1 | Resolving vertically (three terms) |
| $250\times0.96 + R = 48\times9.8$ | | |
| Normal reaction is 230.4 N | A1 | |
## Part (a)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $T\sin\alpha = mg$ | M1 | Vertical equation with $R=0$ |
| $T\times0.96 = 48\times9.8$ | A1 | Or $T\sin73.7=\ldots$ or $T\cos16.3=\ldots$ |
| $T = 490$ | | |
| $490\times0.28 = 48\times\frac{V^2}{8.4}$ | M1 | Obtaining equation for $V$ |
| $V = 4.9$ | A1 | Allow $T=490$ obtained in (i) and used correctly in (ii) for full marks |
## Part (b)(i)
| Answer | Mark | Guidance |
|--------|------|----------|
| $\frac{1}{2}m(v^2-u^2) = m\times9.8(2.5-2.5\cos\theta)$ | M1 | Equation involving KE and PE |
| $v^2 - u^2 = 49(1-\cos\theta)$ | A1 | |
| $v^2 = u^2 + 49 - 49\cos\theta$ | E1 | |
## Part (b)(ii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $mg\cos\theta - R = m\frac{v^2}{r}$ | M1, A1 | Radial equation (three terms) |
| $48\times9.8\left(\frac{u^2+49-v^2}{49}\right) - R = \frac{48v^2}{2.5}$ | M1 | Obtaining equation in $R$, $u$, $v$ |
| $9.6u^2 + 470.4 - 9.6v^2 - R = 19.2v^2$ | | |
| $R = 470.4 + 9.6u^2 - 28.8v^2$ | A1 | |
## Part (b)(iii)
| Answer | Mark | Guidance |
|--------|------|----------|
| $470.4 + 9.6u^2 - 28.8\times4.15^2 = 0$ | M1 | Substituting $R=0$ and $v=4.15$ or other complete method leading to equation for $u$ |
| $u = 1.63$ (3 sf) | A1 | ft requires $0 < u < 4.15$ |
---2
\begin{enumerate}[label=(\alph*)]
\item A particle P , of mass 48 kg , is moving in a horizontal circle of radius 8.4 m at a constant speed of $V \mathrm {~m} \mathrm {~s} ^ { - 1 }$, in contact with a smooth horizontal surface. A light inextensible rope of length 30 m connects P to a fixed point A which is vertically above the centre C of the circle, as shown in Fig. 2.1.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f2dd5719-bef3-45f2-afd2-c481e6a4b129-3_526_490_482_870}
\captionsetup{labelformat=empty}
\caption{Fig. 2.1}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Given that $V = 3.5$, find the tension in the rope and the normal reaction of the surface on P .
\item Calculate the value of $V$ for which the normal reaction is zero.
\end{enumerate}\item The particle P , of mass 48 kg , is now placed on the highest point of a fixed solid sphere with centre O and radius 2.5 m . The surface of the sphere is smooth. The particle P is given an initial horizontal velocity of $u \mathrm {~m} \mathrm {~s} ^ { - 1 }$, and it then moves in part of a vertical circle with centre O and radius 2.5 m . When OP makes an angle $\theta$ with the upward vertical and P is still in contact with the surface of the sphere, P has speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ and the normal reaction of the sphere on P is $R \mathrm {~N}$, as shown in Fig. 2.2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{f2dd5719-bef3-45f2-afd2-c481e6a4b129-3_590_617_1706_804}
\captionsetup{labelformat=empty}
\caption{Fig. 2.2}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Show that $v ^ { 2 } = u ^ { 2 } + 49 - 49 \cos \theta$.
\item Find an expression for $R$ in terms of $u$ and $v$.
\item Given that P loses contact with the surface of the sphere at the instant when its speed is $4.15 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, find the value of $u$.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI M3 2011 Q2 [18]}}