| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2006 |
| Session | January |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: string becomes slack |
| Difficulty | Standard +0.3 This is a standard M3 vertical circle problem requiring energy conservation to find v², resolving forces for tension, and finding when string becomes slack (T=0). While it involves multiple steps and careful angle work, it follows a well-established template that M3 students practice extensively. The 'show that' part guides students through the key relationship, making subsequent parts more routine. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05e Radial/tangential acceleration6.05f Vertical circle: motion including free fall |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| (i) \(0.5\times0.3v^2 - 0.5\times0.3\times2^2 = 0.3\times9.8\times0.5\cos60° - 0.3\times9.8\times0.5\cos\theta\) | M1 | For use of conservation of energy |
| A2,1,0 | \(-1\) each error | |
| \(v^2 = 8.9 - 9.8\cos\theta\) | A1 | [4] AG |
| (ii) \(T + 0.3\times9.8\cos\theta = 0.3v^2/0.5\) | M1, A1 | For using Newton's 2nd law radially |
| \(T + 2.94\cos\theta = 0.6(8.9 - 9.8\cos\theta)\) | M1 | For correct substitution for \(v^2\) |
| Tension is \((5.34 - 8.82\cos\theta)\) N | A1 | [4] Accept any correct form |
| (iii) For using \(T = 0\) | M1 | |
| Basic value \(\theta = 52.7°\) | A1 ft | ft any \(T\) of the form \(a - b\cos\theta\) |
| Angle \(= (360 - 52.7) - 60\) | M1 | |
| Angle turned through is \(247°\) | A1 | [4] |
## Question 6:
| Answer/Working | Marks | Guidance |
|---|---|---|
| **(i)** $0.5\times0.3v^2 - 0.5\times0.3\times2^2 = 0.3\times9.8\times0.5\cos60° - 0.3\times9.8\times0.5\cos\theta$ | M1 | For use of conservation of energy |
| | A2,1,0 | $-1$ each error |
| $v^2 = 8.9 - 9.8\cos\theta$ | A1 | [4] AG |
| **(ii)** $T + 0.3\times9.8\cos\theta = 0.3v^2/0.5$ | M1, A1 | For using Newton's 2nd law radially |
| $T + 2.94\cos\theta = 0.6(8.9 - 9.8\cos\theta)$ | M1 | For correct substitution for $v^2$ |
| Tension is $(5.34 - 8.82\cos\theta)$ N | A1 | [4] Accept any correct form |
| **(iii)** For using $T = 0$ | M1 | |
| Basic value $\theta = 52.7°$ | A1 ft | ft any $T$ of the form $a - b\cos\theta$ |
| Angle $= (360 - 52.7) - 60$ | M1 | |
| Angle turned through is $247°$ | A1 | [4] |
---6\\
\includegraphics[max width=\textwidth, alt={}, center]{5bb3bd29-a2eb-4124-802c-fb17b68c50e4-3_598_839_1480_706}
One end of a light inextensible string of length 0.5 m is attached to a fixed point $O$. A particle $P$ of mass 0.3 kg is attached to the other end of the string. With the string taut and at an angle of $60 ^ { \circ }$ to the upward vertical, $P$ is projected with speed $2 \mathrm {~ms} ^ { - 1 }$ (see diagram). $P$ begins to move without air resistance in a vertical circle with centre $O$. When the string makes an angle $\theta$ with the upward vertical, the speed of $P$ is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$.\\
(i) Show that $v ^ { 2 } = 8.9 - 9.8 \cos \theta$.\\
(ii) Find the tension in the string in terms of $\theta$.\\
(iii) $P$ does not move in a complete circle. Calculate the angle through which $O P$ turns before $P$ leaves the circular path.
\hfill \mbox{\textit{OCR M3 2006 Q6 [12]}}