| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2009 |
| Session | January |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: string becomes slack |
| Difficulty | Standard +0.8 This is a multi-part vertical circle problem requiring energy conservation, resolution of forces in radial/tangential directions, and finding when tension becomes zero. While it involves several standard M3 techniques (energy methods, circular motion dynamics, Newton's second law in polar coordinates), the problem is scaffolded with 'show that' parts that guide students through the solution. The mathematical manipulation is straightforward once the physics setup is understood, making this moderately above average difficulty for A-level but not requiring exceptional insight. |
| Spec | 6.05d Variable speed circles: energy methods6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Loss in PE \(= mg(0.5\sin\theta)\) | B1 | |
| \([\frac{1}{2}mv^2 - \frac{1}{2}m \cdot 3^2 = mg(0.5\sin\theta)]\) | M1 | For using KE gain = PE loss (3 terms required) |
| \(v^2 = 9 + 9.8\sin\theta\) | A1 | AG |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(a_r = 18 + 19.6\sin\theta\) | B1 | Using \(a_r = v^2/0.5\) |
| \([ma_t = mg\cos\theta]\) | M1 | For using Newton's second law tangentially |
| \(a_t = 9.8\cos\theta\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \([T - mg\sin\theta = ma_r]\) | M1 | For using Newton's second law radially (3 terms required) |
| \(T - 1.96\sin\theta = 0.2(18 + 19.6\sin\theta)\) | A1 | |
| \(T = 3.6 + 5.88\sin\theta\) | A1 | AG |
| \(\theta = 3.8\) | B1 |
## Question 4:
**Part (i)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Loss in PE $= mg(0.5\sin\theta)$ | B1 | |
| $[\frac{1}{2}mv^2 - \frac{1}{2}m \cdot 3^2 = mg(0.5\sin\theta)]$ | M1 | For using KE gain = PE loss (3 terms required) |
| $v^2 = 9 + 9.8\sin\theta$ | A1 | AG |
**Part (ii)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $a_r = 18 + 19.6\sin\theta$ | B1 | Using $a_r = v^2/0.5$ |
| $[ma_t = mg\cos\theta]$ | M1 | For using Newton's second law tangentially |
| $a_t = 9.8\cos\theta$ | A1 | |
**Part (iii)**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $[T - mg\sin\theta = ma_r]$ | M1 | For using Newton's second law radially (3 terms required) |
| $T - 1.96\sin\theta = 0.2(18 + 19.6\sin\theta)$ | A1 | |
| $T = 3.6 + 5.88\sin\theta$ | A1 | AG |
| $\theta = 3.8$ | B1 | |
---(i) Show that $v ^ { 2 } = 9 + 9.8 \sin \theta$.\\
(ii) Find, in terms of $\theta$, the radial and tangential components of the acceleration of $P$.\\
(iii) Show that the tension in the string is $( 3.6 + 5.88 \sin \theta ) \mathrm { N }$ and hence find the value of $\theta$ at the instant when the string becomes slack, giving your answer correct to 1 decimal place.
\hfill \mbox{\textit{OCR M3 2009 Q4 [10]}}