| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2011 |
| Session | June |
| Marks | 18 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: tension at specific point |
| Difficulty | Standard +0.3 This is a standard M3 vertical circle problem with routine application of circular motion equations (T + mg = mv²/r at top, resolving forces at angle, energy conservation). Part (b) involves conical pendulum with straightforward force resolution. All techniques are textbook exercises requiring careful calculation but no novel insight—slightly easier than average A-level due to methodical step-by-step structure. |
| Spec | 3.03d Newton's second law: 2D vectors6.05a Angular velocity: definitions6.05c Horizontal circles: conical pendulum, banked tracks6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(0.6 + 0.2\times 9.8 = 0.2\times\frac{u^2}{3.2}\) | M1 A1 | For acceleration \(\frac{u^2}{3.2}\) |
| Speed is \(6.4\) ms\(^{-1}\) | A1 | |
| Total | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{2}m(v^2 - u^2) = m\times 9.8(3.2 + 3.2\cos 60°)\) | M1 A1 | Equation involving KE and PE |
| \(v^2 = 135.04\) | ||
| Radial component is \(\frac{v^2}{3.2} = 42.2\) ms\(^{-2}\) | A1 | (ft is \(29.4 + \frac{u^2}{3.2}\)) |
| Tangential component is \(g\sin 60°\) | M1 | M1A0 for \(g\cos 60°\) |
| \(= 8.49\) ms\(^{-2}\) (3 sf) | A1 | M0 for \(mg\sin 60°\); *If radial and tangential components are interchanged, withhold first A1* |
| Total (A) | 5 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(T - mg\cos 60° = ma\) | M1 | Radial equation (three terms); Allow M1 for \(T - mg = ma\); *This M1 can be awarded in (A)* |
| \(T - 0.2\times 9.8\cos 60° = 0.2\times 42.2\) | A1 | ft dependent on M1 for energy in (A) |
| Tension is \(9.42\) N | A1 cao | *SC* If \(60°\) with upward vertical: (A) M1A0A0 M1A1 (8.49); (B) M1A1A1 (3.54) |
| Total (B) | 3 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(T\cos 36° + 0.75\sin 36° = 0.2\times 9.8\) | M1 | Resolving vertically (three terms); Allow sin/cos confusion, but both \(T\) and \(R\) must be resolved |
| Tension is \(1.88\) N (3 sf) | A1 | |
| Total | 2 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Angular speed \(\omega = \frac{2\pi}{1.8}\) \((= 3.491)\) | B1 | Or \(v = \frac{2\pi r}{1.8}\) |
| \(T\sin 36° - 0.75\cos 36° = 0.2r\left(\frac{2\pi}{1.8}\right)^2\) | M1 A1 | Horiz eqn involving \(r\omega^2\) or \(v^2/r\); Equation for \(r\) (or \(l\)) |
| \(r = 0.204\) | ||
| Length of string is \(\frac{r}{\sin 36°}\) | M1 | *Dependent on previous M1* |
| \(= 0.347\) m (3 sf) | A1 cao | |
| Total | 5 |
# Question 2:
## Part (a)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $0.6 + 0.2\times 9.8 = 0.2\times\frac{u^2}{3.2}$ | M1 A1 | For acceleration $\frac{u^2}{3.2}$ |
| Speed is $6.4$ ms$^{-1}$ | A1 | |
| **Total** | **3** | |
## Part (a)(ii) — Section A
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}m(v^2 - u^2) = m\times 9.8(3.2 + 3.2\cos 60°)$ | M1 A1 | Equation involving KE and PE |
| $v^2 = 135.04$ | | |
| Radial component is $\frac{v^2}{3.2} = 42.2$ ms$^{-2}$ | A1 | (ft is $29.4 + \frac{u^2}{3.2}$) |
| Tangential component is $g\sin 60°$ | M1 | M1A0 for $g\cos 60°$ |
| $= 8.49$ ms$^{-2}$ (3 sf) | A1 | M0 for $mg\sin 60°$; *If radial and tangential components are interchanged, withhold first A1* |
| **Total (A)** | **5** | |
## Part (a)(ii) — Section B
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T - mg\cos 60° = ma$ | M1 | Radial equation (three terms); Allow M1 for $T - mg = ma$; *This M1 can be awarded in (A)* |
| $T - 0.2\times 9.8\cos 60° = 0.2\times 42.2$ | A1 | ft dependent on M1 for energy in (A) |
| Tension is $9.42$ N | A1 cao | *SC* If $60°$ with upward vertical: (A) M1A0A0 M1A1 (8.49); (B) M1A1A1 (3.54) |
| **Total (B)** | **3** | |
## Part (b)(i)
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T\cos 36° + 0.75\sin 36° = 0.2\times 9.8$ | M1 | Resolving vertically (three terms); Allow sin/cos confusion, but both $T$ and $R$ must be resolved |
| Tension is $1.88$ N (3 sf) | A1 | |
| **Total** | **2** | |
## Part (b)(ii)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Angular speed $\omega = \frac{2\pi}{1.8}$ $(= 3.491)$ | B1 | Or $v = \frac{2\pi r}{1.8}$ |
| $T\sin 36° - 0.75\cos 36° = 0.2r\left(\frac{2\pi}{1.8}\right)^2$ | M1 A1 | Horiz eqn involving $r\omega^2$ or $v^2/r$; Equation for $r$ (or $l$) |
| $r = 0.204$ | | |
| Length of string is $\frac{r}{\sin 36°}$ | M1 | *Dependent on previous M1* |
| $= 0.347$ m (3 sf) | A1 cao | |
| **Total** | **5** | |
---2
\begin{enumerate}[label=(\alph*)]
\item A particle P of mass 0.2 kg is connected to a fixed point O by a light inextensible string of length 3.2 m , and is moving in a vertical circle with centre O and radius 3.2 m . Air resistance may be neglected. When P is at the highest point of the circle, the tension in the string is 0.6 N .
\begin{enumerate}[label=(\roman*)]
\item Find the speed of P when it is at the highest point.
\item For an instant when OP makes an angle of $60 ^ { \circ }$ with the downward vertical, find\\
(A) the radial and tangential components of the acceleration of P ,\\
(B) the tension in the string.
\end{enumerate}\item A solid cone is fixed with its axis of symmetry vertical and its vertex V uppermost. The semivertical angle of the cone is $36 ^ { \circ }$, and its surface is smooth. A particle Q of mass 0.2 kg is connected to V by a light inextensible string, and Q moves in a horizontal circle at constant speed, in contact with the surface of the cone, as shown in Fig. 2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{5ecb198d-7863-4fc2-81b6-c8b6c37b1859-3_455_609_950_808}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
The particle Q makes one complete revolution in 1.8 s , and the normal reaction of the cone on Q has magnitude 0.75 N .
\begin{enumerate}[label=(\roman*)]
\item Find the tension in the string.
\item Find the length of the string.
\end{enumerate}\end{enumerate}
\hfill \mbox{\textit{OCR MEI M3 2011 Q2 [18]}}