| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2012 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: complete revolution conditions |
| Difficulty | Standard +0.3 This is a standard two-part circular motion question covering banked curves and vertical circles. Part (a) involves routine resolution of forces and applying F=ma for circular motion with straightforward trigonometry. Part (b) requires energy conservation and the standard condition for string becoming slack (tension = 0), both being textbook applications. While multi-step, all techniques are standard M3 material with no novel insight required, making it slightly easier than average. |
| Spec | 6.05c Horizontal circles: conical pendulum, banked tracks6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(\frac{1}{2}m(7^2 - 2.8^2) = mg(a + a\cos\theta)\) | M1 | Equation involving KE and PE; \(h = 2.1\) implies M1, \(a\) is length of string |
| Correct equation involving \(a\) and \(\theta\); \(a(1 + \cos\theta) = 2.1\) | A1 | Might use angle with downward vertical or horizontal |
| \(mg\cos\theta = m \times \frac{2.8^2}{a}\); \(a\cos\theta = 0.8\) | M1, A1 | Radial equation of motion; correct equation involving \(a\) and \(\theta\); might also involve \(T\) |
| Eliminating \(\theta\) or \(a\) | M1 | Dependent on previous M1M1 |
| Length of string is \(1.3\) m | A1 | |
| Angle with upward vertical is \(52.0°\) (3 sf) | A1 [7] | A0 for \(128°\) or \(38°\) |
## Question 2(b):
| Answer | Marks | Guidance |
|--------|-------|----------|
| $\frac{1}{2}m(7^2 - 2.8^2) = mg(a + a\cos\theta)$ | M1 | Equation involving KE and PE; $h = 2.1$ implies M1, $a$ is length of string |
| Correct equation involving $a$ and $\theta$; $a(1 + \cos\theta) = 2.1$ | A1 | Might use angle with downward vertical or horizontal |
| $mg\cos\theta = m \times \frac{2.8^2}{a}$; $a\cos\theta = 0.8$ | M1, A1 | Radial equation of motion; correct equation involving $a$ and $\theta$; might also involve $T$ |
| Eliminating $\theta$ or $a$ | M1 | Dependent on previous M1M1 |
| Length of string is $1.3$ m | A1 | |
| Angle with upward vertical is $52.0°$ (3 sf) | A1 [7] | A0 for $128°$ or $38°$ |
---2
\begin{enumerate}[label=(\alph*)]
\item Fig. 2 shows a car of mass 800 kg moving at constant speed in a horizontal circle with centre C and radius 45 m , on a road which is banked at an angle of $18 ^ { \circ }$ to the horizontal. The forces shown are the weight $W$ of the car, the normal reaction, $R$, of the road on the car and the frictional force $F$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{86dd0c01-970d-4b67-9a6c-5df276a4a2be-3_286_970_402_561}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Given that the frictional force is zero, find the speed of the car.
\item Given instead that the speed of the car is $15 \mathrm {~m} \mathrm {~s} ^ { - 1 }$, find the frictional force and the normal reaction.
\end{enumerate}\item One end of a light inextensible string is attached to a fixed point O , and the other end is attached to a particle P of mass $m \mathrm {~kg}$. Starting with the string taut and P vertically below $\mathrm { O } , \mathrm { P }$ is set in motion with a horizontal velocity of $7 \mathrm {~ms} ^ { - 1 }$. It then moves in part of a vertical circle with centre O . The string becomes slack when the speed of P is $2.8 \mathrm {~ms} ^ { - 1 }$.
Find the length of the string. Find also the angle that OP makes with the upward vertical at the instant when the string becomes slack.
\end{enumerate}
\hfill \mbox{\textit{OCR MEI M3 2012 Q2 [18]}}