| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2008 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: string becomes slack |
| Difficulty | Standard +0.8 This is a classic vertical circle problem requiring energy conservation and circular motion dynamics. Students must find speed at horizontal position using energy methods, calculate tension using centripetal force, then determine the slack condition (T=0) which requires solving a transcendental equation involving both energy and force equations. The multi-step nature and need to combine two principles (energy + Newton's 2nd law) elevates this above routine questions, though it follows a standard M3 template. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \([\frac{1}{2}mv_7^2 = \frac{1}{2}mv^2 + 2\text{mg}]\) | M1 A1 M1 | For using the principle of conservation of energy; For using Newton's second law horizontally and \(a = V/r\); For using Newton's second law radially |
| Speed is 3.13ms⁻¹ | A1 | |
| \([T = mv^2/r]\) | ||
| Tension is 1.96N | A1ft | For using Newton's second law radially; It with \(T = 0\) (may be implied) |
| (ii) \(v^2 = -2g\cos\theta\) | A1 M1 | For using the principle of conservation of energy |
| \(\frac{1}{2}mv_7^2 = \frac{1}{2}mv^2 + \text{mg}(2 - 2\cos\theta)\) | ||
| \([-2g\cos\theta = 49 - 4g + 4g\cos\theta]\) | A1 M1 | For eliminating \(v^2\); May be implied by answer |
| \(6g\cos\theta = -9.8\) | A1 | |
| \(\theta = 99.6\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| (ii) \([T - \text{mg}\cos\theta = mv^2/r]\) | M1 M1 | For using Newton's second law radially or using the principle of conservation of energy |
| \(\frac{1}{2}mv_7^2 = \frac{1}{2}mv^2 + \text{mg}(2 - 2\cos\theta)\) | ||
| \([T - \text{mg}\cos\theta = m(49 - 4g + 4g\cos\theta)/2]\) | A1 M1 | For eliminating \(v^2\); For using \(T = 0\) (may be implied) It error in energy equation |
| \(-2g\cos\theta = 49 - 4g + 4g\cos\theta\) | A1ft | May be implied by answer |
| \(6g\cos\theta = -9.8\) | A1 | |
| \(\theta = 99.6\) | A1 |
**(i)** $[\frac{1}{2}mv_7^2 = \frac{1}{2}mv^2 + 2\text{mg}]$ | M1 A1 M1 | For using the principle of conservation of energy; For using Newton's second law horizontally and $a = V/r$; For using Newton's second law radially
Speed is 3.13ms⁻¹ | A1 |
$[T = mv^2/r]$ | |
Tension is 1.96N | A1ft | For using Newton's second law radially; It with $T = 0$ (may be implied)
**(ii)** $v^2 = -2g\cos\theta$ | A1 M1 | For using the principle of conservation of energy
$\frac{1}{2}mv_7^2 = \frac{1}{2}mv^2 + \text{mg}(2 - 2\cos\theta)$ | |
$[-2g\cos\theta = 49 - 4g + 4g\cos\theta]$ | A1 M1 | For eliminating $v^2$; May be implied by answer
$6g\cos\theta = -9.8$ | A1 |
$\theta = 99.6$ | A1 |
**Alternative for candidates who eliminate $v^2$ before using $T = 0$:**
**(ii)** $[T - \text{mg}\cos\theta = mv^2/r]$ | M1 M1 | For using Newton's second law radially or using the principle of conservation of energy
$\frac{1}{2}mv_7^2 = \frac{1}{2}mv^2 + \text{mg}(2 - 2\cos\theta)$ | |
$[T - \text{mg}\cos\theta = m(49 - 4g + 4g\cos\theta)/2]$ | A1 M1 | For eliminating $v^2$; For using $T = 0$ (may be implied) It error in energy equation
$-2g\cos\theta = 49 - 4g + 4g\cos\theta$ | A1ft | May be implied by answer
$6g\cos\theta = -9.8$ | A1 |
$\theta = 99.6$ | A1 |
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{7e0f600a-18f1-458b-8549-27fca592b19c-4_497_524_276_804}
A particle $P$ of mass 0.4 kg is attached to one end of a light inextensible string of length 2 m . The other end of the string is attached to a fixed point $O$. With the string taut the particle is travelling in a circular path in a vertical plane. The angle between the string and the downward vertical is $\theta ^ { \circ }$ (see diagram). When $\theta = 0$ the speed of $P$ is $7 \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) At the instant when the string is horizontal, find the speed of $P$ and the tension in the string.\\
(ii) At the instant when the string becomes slack, find the value of $\theta$.
\hfill \mbox{\textit{OCR M3 2008 Q6 [12]}}