| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2014 |
| Session | June |
| 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, requiring basic trigonometry and centripetal force; (b) applies standard vertical circle formulas with energy conservation. All techniques are textbook exercises with no novel problem-solving required, making it slightly easier than average for M3 level. |
| Spec | 6.05c Horizontal circles: conical pendulum, banked tracks6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(T\cos\alpha = T\cos\beta + 0.24 \times 9.8\) | M1 | Resolving vertically (three terms); \(\alpha = \hat{A} = 28.1°\), \(\beta = \hat{B} = 53.1°\) |
| \(\frac{15}{17}T = \frac{3}{5}T + 2.352\) | A1 | Accept \(\cos 28.1°\) etc |
| Tension is \(8.33\) N | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \(T\sin\alpha + T\sin\beta = m(r\omega^2)\) | M1 | Eqn with resolved tension and \(r\omega^2\); one tension sufficient for M1; allow \(\frac{v^2}{r}\) for M1 |
| \(\frac{8}{17}T + \frac{4}{5}T = (0.24)(1.6\omega^2)\) | A1 | |
| Angular speed is \(5.25\) rad s\(^{-1}\) | A1 [3] | FT is \(1.819\sqrt{T}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Equation with tension, resolved weight and \(v_1^2/r\) | M1 | Accept use of mass instead of weight throughout for M marks |
| \(15 - (0.3)(9.8)\cos 25° = (0.3)\dfrac{v_1^2}{1.8}\) | A1 | |
| Speed is \(8.60\) ms\(^{-1}\) (3 sf) | A1 [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Equation with initial KE, final KE and attempt at PE | M1 | |
| \(\frac{1}{2}m(v_1^2 - v_2^2) = mg(1.8\cos 25° + 1.8\cos 60°)\) | A1 | |
| \(v_2^2 = 24.40\) | ||
| Equation with tension, resolved weight (using \(60°\)) and \(v_2^2/r\) | M1 | |
| \(T + (0.3)(9.8)\cos 60° = (0.3)\dfrac{v_2^2}{1.8}\) | A1 | SC For \(60°\) with downward vertical give A1 for \(11.4\) N (after M1A0M1A0), i.e. 3/5 |
| Tension is \(2.60\) N (3 sf) | A1 [5] | FT is \(\frac{v_1^2}{6} - 9.739\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| \((m)g\cos\theta = (m)\dfrac{v_3^2}{1.8}\) | M1, A1 | Equation with resolved weight in general position and \(v_3^2/r\); \(\theta\) is angle between OP and upward vertical; may also include \(T\) |
| \(\frac{1}{2}m(v_2^2 - v_3^2) = mg(1.8)(\cos\theta - \cos 60°)\) | M1, A1 | Equation with KE and attempt at PE in general position; OR \(\frac{1}{2}m(v_1^2-v_3^2)=mg(1.8)(\cos\theta+\cos 25°)\) |
| \(24.40 - v_3^2 = 2v_3^2 - 9.8 \times 1.8\) | \(\cos\theta = 0.794\), \(\theta = 0.653\) rad \(= 37.4°\) | |
| Speed is \(3.74\) ms\(^{-1}\) (3 sf) | A1 [5] | CAO |
# Question 2:
## Part (a)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| $T\cos\alpha = T\cos\beta + 0.24 \times 9.8$ | M1 | Resolving vertically (three terms); $\alpha = \hat{A} = 28.1°$, $\beta = \hat{B} = 53.1°$ |
| $\frac{15}{17}T = \frac{3}{5}T + 2.352$ | A1 | Accept $\cos 28.1°$ etc |
| Tension is $8.33$ N | A1 [3] | |
## Part (a)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $T\sin\alpha + T\sin\beta = m(r\omega^2)$ | M1 | Eqn with resolved tension and $r\omega^2$; one tension sufficient for M1; allow $\frac{v^2}{r}$ for M1 |
| $\frac{8}{17}T + \frac{4}{5}T = (0.24)(1.6\omega^2)$ | A1 | |
| Angular speed is $5.25$ rad s$^{-1}$ | A1 [3] | FT is $1.819\sqrt{T}$ |
## Part (b)(i):
| Answer | Mark | Guidance |
|--------|------|----------|
| Equation with tension, resolved weight and $v_1^2/r$ | M1 | Accept use of mass instead of weight throughout for M marks |
| $15 - (0.3)(9.8)\cos 25° = (0.3)\dfrac{v_1^2}{1.8}$ | A1 | |
| Speed is $8.60$ ms$^{-1}$ (3 sf) | A1 [3] | |
## Part (b)(ii):
| Answer | Mark | Guidance |
|--------|------|----------|
| Equation with initial KE, final KE and attempt at PE | M1 | |
| $\frac{1}{2}m(v_1^2 - v_2^2) = mg(1.8\cos 25° + 1.8\cos 60°)$ | A1 | |
| $v_2^2 = 24.40$ | | |
| Equation with tension, resolved weight (using $60°$) and $v_2^2/r$ | M1 | |
| $T + (0.3)(9.8)\cos 60° = (0.3)\dfrac{v_2^2}{1.8}$ | A1 | SC For $60°$ with downward vertical give A1 for $11.4$ N (after M1A0M1A0), i.e. 3/5 |
| Tension is $2.60$ N (3 sf) | A1 [5] | FT is $\frac{v_1^2}{6} - 9.739$ |
## Part (b)(iii):
| Answer | Mark | Guidance |
|--------|------|----------|
| $(m)g\cos\theta = (m)\dfrac{v_3^2}{1.8}$ | M1, A1 | Equation with resolved weight in general position and $v_3^2/r$; $\theta$ is angle between OP and upward vertical; may also include $T$ |
| $\frac{1}{2}m(v_2^2 - v_3^2) = mg(1.8)(\cos\theta - \cos 60°)$ | M1, A1 | Equation with KE and attempt at PE in general position; OR $\frac{1}{2}m(v_1^2-v_3^2)=mg(1.8)(\cos\theta+\cos 25°)$ |
| $24.40 - v_3^2 = 2v_3^2 - 9.8 \times 1.8$ | | $\cos\theta = 0.794$, $\theta = 0.653$ rad $= 37.4°$ |
| Speed is $3.74$ ms$^{-1}$ (3 sf) | A1 [5] | CAO |
---2
\begin{enumerate}[label=(\alph*)]
\item The fixed point A is vertically above the fixed point B . A light inextensible string of length 5.4 m has one end attached to A and the other end attached to B. The string passes through a small smooth ring R of mass 0.24 kg , and R is moving at constant angular speed in a horizontal circle. The circle has radius 1.6 m , and $\mathrm { AR } = 3.4 \mathrm {~m} , \mathrm { RB } = 2.0 \mathrm {~m}$, as shown in Fig. 2 .
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5a0df44f-f8f0-44d4-b2f6-70a5314706f9-3_565_504_447_753}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
\begin{enumerate}[label=(\roman*)]
\item Find the tension in the string.
\item Find the angular speed of R .
\end{enumerate}\item A particle P of mass 0.3 kg is joined to a fixed point O by a light inextensible string of length 1.8 m . The particle P moves without resistance in part of a vertical circle with centre O and radius 1.8 m . When OP makes an angle of $25 ^ { \circ }$ with the downward vertical, the tension in the string is 15 N .
\begin{enumerate}[label=(\roman*)]
\item Find the speed of P when OP makes an angle of $25 ^ { \circ }$ with the downward vertical.
\item Find the tension in the string when OP makes an angle of $60 ^ { \circ }$ with the upward vertical.
\item Find the speed of P at the instant when the string becomes slack.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI M3 2014 Q2 [19]}}