| Exam Board | OCR MEI |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2008 |
| Session | June |
| 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 straightforward application of T cos θ = mg, T sin θ = mv²/r for the conical pendulum, then energy conservation and F = mv²/r for vertical circle motion. All parts follow routine procedures with no novel problem-solving required, making it slightly easier than average for Further Maths M3 content. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05a Angular velocity: definitions6.05c Horizontal circles: conical pendulum, banked tracks6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T\cos\alpha = mg\) | M1 | Resolving vertically *(condone sin/cos mix for M marks throughout this question)* |
| \(3.92\cos\alpha = 0.3 \times 9.8\) | ||
| \(\cos\alpha = 0.75\) | ||
| Angle is \(41.4°\) (0.723 rad) | A1 | 2 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T\sin\alpha = m\frac{v^2}{r}\) | M1 | Force and acceleration towards centre *(condone \(v^2/4.2\) or \(4.2\omega^2\))* |
| \(3.92\sin\alpha = 0.3 \times \frac{v^2}{4.2\sin\alpha}\) | B1 | For radius is \(4.2\sin\alpha\ (=2.778)\) |
| Speed is \(4.9\ \text{ms}^{-1}\) | A1 | Not awarded for equation in \(\omega\) unless \(v = (4.2\sin\alpha)\omega\) also appears |
| A1 | 4 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(T - mg\cos\theta = m\frac{v^2}{a}\) | M1 | Forces and acceleration towards O |
| \(T - 0.3 \times 9.8 \times \cos 60° = 0.3 \times \frac{8.4^2}{4.2}\) | A1 | |
| Tension is \(6.51\) N | A1 | 3 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2}mv^2 - mg \times 4.2\cos\theta = \frac{1}{2}m \times 8.4^2 - mg \times 4.2\cos 60°\) | M1, M1, A1 | For \((-) mg \times 4.2\cos\theta\) in PE; equation involving \(\frac{1}{2}mv^2\) and PE |
| \(v^2 - 82.32\cos\theta = 70.56 - 41.16\) | E1 | |
| \(v^2 = 29.4 + 82.32\cos\theta\) | 4 marks total |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \((T) - mg\cos\theta = m\frac{v^2}{a}\) | M1 | Force and acceleration towards O |
| \((T) - m \times 9.8\cos\theta = m \times \frac{29.4 + 82.32\cos\theta}{4.2}\) | M1, A1 | Substituting for \(v^2\) |
| String becomes slack when \(T = 0\) | M1 | Dependent on first M1 |
| \(-9.8\cos\theta = 7 + 19.6\cos\theta\) | ||
| \(\cos\theta = -\frac{7}{29.4}\) | ||
| \(\theta = 104°\) (1.81 rad) | A1 | *No marks for \(v=0 \Rightarrow \theta = 111°\)*; 5 marks total |
# Question 2:
## Part (i):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T\cos\alpha = mg$ | M1 | Resolving vertically *(condone sin/cos mix for M marks throughout this question)* |
| $3.92\cos\alpha = 0.3 \times 9.8$ | | |
| $\cos\alpha = 0.75$ | | |
| Angle is $41.4°$ (0.723 rad) | A1 | **2 marks total** |
## Part (ii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T\sin\alpha = m\frac{v^2}{r}$ | M1 | Force and acceleration towards centre *(condone $v^2/4.2$ or $4.2\omega^2$)* |
| $3.92\sin\alpha = 0.3 \times \frac{v^2}{4.2\sin\alpha}$ | B1 | For radius is $4.2\sin\alpha\ (=2.778)$ |
| Speed is $4.9\ \text{ms}^{-1}$ | A1 | Not awarded for equation in $\omega$ unless $v = (4.2\sin\alpha)\omega$ also appears |
| | A1 | **4 marks total** |
## Part (iii):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $T - mg\cos\theta = m\frac{v^2}{a}$ | M1 | Forces and acceleration towards O |
| $T - 0.3 \times 9.8 \times \cos 60° = 0.3 \times \frac{8.4^2}{4.2}$ | A1 | |
| Tension is $6.51$ N | A1 | **3 marks total** |
## Part (iv):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}mv^2 - mg \times 4.2\cos\theta = \frac{1}{2}m \times 8.4^2 - mg \times 4.2\cos 60°$ | M1, M1, A1 | For $(-) mg \times 4.2\cos\theta$ in PE; equation involving $\frac{1}{2}mv^2$ and PE |
| $v^2 - 82.32\cos\theta = 70.56 - 41.16$ | E1 | |
| $v^2 = 29.4 + 82.32\cos\theta$ | | **4 marks total** |
## Part (v):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $(T) - mg\cos\theta = m\frac{v^2}{a}$ | M1 | Force and acceleration towards O |
| $(T) - m \times 9.8\cos\theta = m \times \frac{29.4 + 82.32\cos\theta}{4.2}$ | M1, A1 | Substituting for $v^2$ |
| String becomes slack when $T = 0$ | M1 | Dependent on first M1 |
| $-9.8\cos\theta = 7 + 19.6\cos\theta$ | | |
| $\cos\theta = -\frac{7}{29.4}$ | | |
| $\theta = 104°$ (1.81 rad) | A1 | *No marks for $v=0 \Rightarrow \theta = 111°$*; **5 marks total** |
---2 A particle P of mass 0.3 kg is connected to a fixed point O by a light inextensible string of length 4.2 m .
Firstly, P is moving in a horizontal circle as a conical pendulum, with the string making a constant angle with the vertical. The tension in the string is 3.92 N .\\
(i) Find the angle which the string makes with the vertical.\\
(ii) Find the speed of P .
P now moves in part of a vertical circle with centre O and radius 4.2 m . When the string makes an angle $\theta$ with the downward vertical, the speed of P is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$ (see Fig. 2). You are given that $v = 8.4$ when $\theta = 60 ^ { \circ }$.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{2a4afead-e772-4d86-bc8d-86ffa5bca507-2_382_648_1985_751}
\captionsetup{labelformat=empty}
\caption{Fig. 2}
\end{center}
\end{figure}
(iii) Find the tension in the string when $\theta = 60 ^ { \circ }$.\\
(iv) Show that $v ^ { 2 } = 29.4 + 82.32 \cos \theta$.\\
(v) Find $\theta$ at the instant when the string becomes slack.
\hfill \mbox{\textit{OCR MEI M3 2008 Q2 [18]}}