| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2009 |
| Session | January |
| Marks | 19 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Two strings/rods system |
| Difficulty | Standard +0.3 This is a standard M3 circular motion question with two routine parts: (a) involves resolving forces on a conical pendulum setup with given geometry, and (b) covers vertical circular motion with standard energy/force equations. All parts use well-practiced techniques (resolving forces, applying F=ma for circular motion, using conservation of energy) with straightforward geometry. The 'show that' parts guide students to expected answers, reducing problem-solving demand. Slightly above average difficulty due to the two-part structure and need for careful resolution of forces in 3D geometry, but remains a textbook exercise. |
| Spec | 6.05c Horizontal circles: conical pendulum, banked tracks6.05d Variable speed circles: energy methods6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(T\cos\alpha = T\cos\beta + 0.27\times9.8\) | M1 | Resolving vertically (weight and at least one resolved tension) |
| \(\sin\alpha = \frac{1.2}{2.0} = \frac{3}{5}\), \(\cos\alpha = \frac{4}{5}\) \((\alpha = 36.87°)\) | A1 | Allow \(T_1\) and \(T_2\) |
| \(\sin\beta = \frac{1.2}{1.3} = \frac{12}{13}\), \(\cos\beta = \frac{5}{13}\) \((\beta = 67.38°)\) | B1 | For \(\cos\alpha\) and \(\cos\beta\) [or \(\alpha\) and \(\beta\)] |
| \(\frac{27}{65}T = 2.646\) | M1 | Obtaining numerical equation for \(T\), e.g. \(T(\cos 36.9 - \cos 67.4) = 0.27\times9.8\) |
| Tension is 6.37 N | E1 | *(Condone 6.36 to 6.38)* |
| Total: 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(T\sin\alpha + T\sin\beta = 0.27\times\frac{v^2}{1.2}\) | M1 | Using \(v^2/1.2\) |
| A1 | Allow \(T_1\) and \(T_2\) | |
| \(6.37\times\frac{3}{5} + 6.37\times\frac{12}{13} = 0.27\times\frac{v^2}{1.2}\) | M1 | Obtaining numerical equation for \(v^2\) |
| \(v^2 = 43.12\) | ||
| Speed is \(6.57\ \text{ms}^{-1}\) | A1 | |
| Total: 4 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(0.2\times9.8 = 0.2\times\frac{u^2}{1.25}\) | M1 | Using acceleration \(u^2/1.25\) |
| \(u^2 = 9.8\times1.25 = 12.25\) | ||
| Speed is \(3.5\ \text{ms}^{-1}\) | E1 | |
| Total: 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{2}m(v^2 - 3.5^2) = mg(1.25 - 1.25\cos60)\) | M1 | Using conservation of energy |
| \(v^2 = 24.5\) | A1 | |
| Radial component is \(\frac{24.5}{1.25} = 19.6\ \text{ms}^{-2}\) | M1 | With numerical value obtained by using energy |
| A1 | (M0 if mass, or another term, included) | |
| Tangential component is \(g\sin60 = 8.49\ \text{ms}^{-2}\) | M1 | For sight of \((m)g\sin60°\) with no other terms |
| A1 | ||
| Total: 6 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(T + 0.2\times9.8\cos60 = 0.2\times19.6\) | M1 | Radial equation (3 terms) |
| Tension is 2.94 N | A1 cao | *This M1 can be awarded in (ii)* |
| Total: 2 |
# Question 2:
## Part (a)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T\cos\alpha = T\cos\beta + 0.27\times9.8$ | M1 | Resolving vertically (weight and at least one resolved tension) |
| $\sin\alpha = \frac{1.2}{2.0} = \frac{3}{5}$, $\cos\alpha = \frac{4}{5}$ $(\alpha = 36.87°)$ | A1 | Allow $T_1$ and $T_2$ |
| $\sin\beta = \frac{1.2}{1.3} = \frac{12}{13}$, $\cos\beta = \frac{5}{13}$ $(\beta = 67.38°)$ | B1 | For $\cos\alpha$ and $\cos\beta$ [or $\alpha$ and $\beta$] |
| $\frac{27}{65}T = 2.646$ | M1 | Obtaining numerical equation for $T$, e.g. $T(\cos 36.9 - \cos 67.4) = 0.27\times9.8$ |
| Tension is 6.37 N | E1 | *(Condone 6.36 to 6.38)* |
| **Total: 5** | | |
## Part (a)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T\sin\alpha + T\sin\beta = 0.27\times\frac{v^2}{1.2}$ | M1 | Using $v^2/1.2$ |
| | A1 | Allow $T_1$ and $T_2$ |
| $6.37\times\frac{3}{5} + 6.37\times\frac{12}{13} = 0.27\times\frac{v^2}{1.2}$ | M1 | Obtaining numerical equation for $v^2$ |
| $v^2 = 43.12$ | | |
| Speed is $6.57\ \text{ms}^{-1}$ | A1 | |
| **Total: 4** | | |
## Part (b)(i):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $0.2\times9.8 = 0.2\times\frac{u^2}{1.25}$ | M1 | Using acceleration $u^2/1.25$ |
| $u^2 = 9.8\times1.25 = 12.25$ | | |
| Speed is $3.5\ \text{ms}^{-1}$ | E1 | |
| **Total: 2** | | |
## Part (b)(ii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}m(v^2 - 3.5^2) = mg(1.25 - 1.25\cos60)$ | M1 | Using conservation of energy |
| $v^2 = 24.5$ | A1 | |
| Radial component is $\frac{24.5}{1.25} = 19.6\ \text{ms}^{-2}$ | M1 | With numerical value obtained by using energy |
| | A1 | (M0 if mass, or another term, included) |
| Tangential component is $g\sin60 = 8.49\ \text{ms}^{-2}$ | M1 | For sight of $(m)g\sin60°$ with no other terms |
| | A1 | |
| **Total: 6** | | |
## Part (b)(iii):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T + 0.2\times9.8\cos60 = 0.2\times19.6$ | M1 | Radial equation (3 terms) |
| Tension is 2.94 N | A1 cao | *This M1 can be awarded in (ii)* |
| **Total: 2** | | |
---2
\begin{enumerate}[label=(\alph*)]
\item Fig. 2 shows a light inextensible string of length 3.3 m passing through a small smooth ring R of mass 0.27 kg . The ends of the string are attached to fixed points A and B , where A is vertically above $B$. The ring $R$ is moving with constant speed in a horizontal circle of radius $1.2 \mathrm {~m} , \mathrm { AR } = 2.0 \mathrm {~m}$ and $\mathrm { BR } = 1.3 \mathrm {~m}$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{b8573ee2-771c-4a93-88d9-346a9da94494-3_570_659_493_781}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Show that the tension in the string is 6.37 N .
\item Find the speed of R .
\end{enumerate}\item One end of a light inextensible string of length 1.25 m is attached to a fixed point O . The other end is attached to a particle P of mass 0.2 kg . The particle P is moving in a vertical circle with centre O and radius 1.25 m , and when P is at the highest point of the circle there is no tension in the string.
\begin{enumerate}[label=(\roman*)]
\item Show that when P is at the highest point its speed is $3.5 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
For the instant when the string OP makes an angle of $60 ^ { \circ }$ with the upward vertical, find
\item the radial and tangential components of the acceleration of P ,
\item the tension in the string.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI M3 2009 Q2 [19]}}