| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2014 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Smooth ring on rotating string |
| Difficulty | Standard +0.8 This is a multi-part circular motion problem requiring resolution of forces in two configurations, geometric reasoning to find string lengths, and application of centripetal force. Parts (i)-(ii) involve standard tension resolution with two different angles, while parts (iii)-(iv) require geometric insight about the conical pendulum configuration and solving simultaneous equations. The multiple steps, geometric complexity, and need to link different configurations make this moderately challenging, though the techniques are standard M2 material. |
| Spec | 6.05c Horizontal circles: conical pendulum, banked tracks6.05d Variable speed circles: energy methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(T\cos30 + T\cos45 = 0.4g\) | M1 | Resolve vertically (3 terms); may be different \(T\)'s at this stage |
| \(T = 2.49\) N | A1 | |
| A1 | \(T = 2.4918\ldots\) | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(cv(T)\sin30 + cv(T)\sin45 = 0.4v^2/0.5\) | M1 | Resolve horizontally (3 terms); may be different \(T\)'s at this stage |
| \(v = 1.94\) m s\(^{-1}\) | A1 | Or use acceleration \(= 0.5\omega^2\) |
| A1 | \(v = 1.93904\ldots\) | |
| [3] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \((2AP =)\dfrac{0.5}{\sin45} + \dfrac{0.5}{\sin30}\) | M1 | Reasonable attempt to use trigonometry to find total length of string |
| \(AP = 0.854\) m | A1 | AG (\(AP = 0.85355\ldots\)m) |
| [2] |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Marks | Guidance |
| \(2T\sin\theta = 0.4(0.854\sin\theta)(3.46^2)\) | M1 | \(\theta\) angle with vertical. Resolve horizontally. Allow with \(T\) only. \(r =\) component of 0.854 |
| \(T = 2.04\) N | A1 | \(T = 2.04474\ldots\) N using \(AP = 0.854\) m, \(T = 2.04367\ldots\) N using exact \(AP\) |
| \(2T\cos\theta = 0.4g\) | M1 | \(\theta\) angle with vertical. Resolve vertically. Allow with \(T\) only |
| \(\theta = 16.5°\) or \(16.6°\) | A1 | \(\theta = 16.55377\ldots°\) using \(AP = 0.854\) m, \(\theta = 16.4526\ldots°\) using exact \(AP\) |
| [4] | SC M1A0M1A1 for use of \(T\) instead of \(2T\) throughout |
## Question 7:
### Part (i)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $T\cos30 + T\cos45 = 0.4g$ | M1 | Resolve vertically (3 terms); may be different $T$'s at this stage |
| $T = 2.49$ N | A1 | |
| | A1 | $T = 2.4918\ldots$ |
| **[3]** | | |
### Part (ii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $cv(T)\sin30 + cv(T)\sin45 = 0.4v^2/0.5$ | M1 | Resolve horizontally (3 terms); may be different $T$'s at this stage |
| $v = 1.94$ m s$^{-1}$ | A1 | Or use acceleration $= 0.5\omega^2$ |
| | A1 | $v = 1.93904\ldots$ |
| **[3]** | | |
### Part (iii)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $(2AP =)\dfrac{0.5}{\sin45} + \dfrac{0.5}{\sin30}$ | M1 | Reasonable attempt to use trigonometry to find total length of string |
| $AP = 0.854$ m | A1 | **AG** ($AP = 0.85355\ldots$m) |
| **[2]** | | |
### Part (iv)
| Answer | Marks | Guidance |
|--------|-------|----------|
| $2T\sin\theta = 0.4(0.854\sin\theta)(3.46^2)$ | M1 | $\theta$ angle with vertical. Resolve horizontally. Allow with $T$ only. $r =$ component of 0.854 |
| $T = 2.04$ N | A1 | $T = 2.04474\ldots$ N using $AP = 0.854$ m, $T = 2.04367\ldots$ N using exact $AP$ |
| $2T\cos\theta = 0.4g$ | M1 | $\theta$ angle with vertical. Resolve vertically. Allow with $T$ only |
| $\theta = 16.5°$ or $16.6°$ | A1 | $\theta = 16.55377\ldots°$ using $AP = 0.854$ m, $\theta = 16.4526\ldots°$ using exact $AP$ |
| **[4]** | | SC M1A0M1A1 for use of $T$ instead of $2T$ throughout |
---7\\
\includegraphics[max width=\textwidth, alt={}, center]{5bfd0285-71cb-4dcb-8545-a379653f9a3e-4_529_403_264_829}
A small smooth ring $P$ of mass 0.4 kg is threaded onto a light inextensible string fixed at $A$ and $B$ as shown in the diagram, with $A$ vertically above $B$. The string is inclined to the vertical at angles of $30 ^ { \circ }$ and $45 ^ { \circ }$ at $A$ and $B$ respectively. $P$ moves in a horizontal circle of radius 0.5 m about a point $C$ vertically below $B$.\\
(i) Calculate the tension in the string.\\
(ii) Calculate the speed of $P$.
The end of the string at $B$ is moved so both ends of the string are now fixed at $A$.\\
(iii) Show that, when the string is taut, $A P$ is now 0.854 m correct to 3 significant figures.\\
$P$ moves in a horizontal circle with angular speed $3.46 \mathrm { rad } \mathrm { s } ^ { - 1 }$.\\
(iv) Find the tension in the string and the angle that the string now makes with the vertical.
\hfill \mbox{\textit{OCR M2 2014 Q7 [12]}}