| Exam Board | OCR |
|---|---|
| Module | M3 (Mechanics 3) |
| Marks | 14 |
| 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, circular motion dynamics, and finding when the string becomes slack. While it involves standard M3 techniques (energy equation, resolving forces radially, setting tension to zero), it requires careful coordinate setup and solving a transcendental equation. The 'show that' part guides students through the energy step, but parts (ii) and (iii) demand independent application of Newton's second law in circular motion and interpretation of the slack condition. This is moderately challenging for M3 level, above average difficulty but not requiring exceptional insight. |
| Spec | 6.02i Conservation of energy: mechanical energy principle6.05e Radial/tangential acceleration6.05f Vertical circle: motion including free fall |
| Answer | Marks | Guidance |
|---|---|---|
| [u sin30° = 3] | M1 | For momentum equation for B, normal to line of centres |
| u = 6 | A1 [2] |
| Answer | Marks | Guidance |
|---|---|---|
| [4sin88.1° = v sin45°] | M1 | For momentum equation for A, normal to line of centres |
| v = 5.65 | A1 M1 | For momentum equation along line of centres |
| 0.4(4cos88.1°) - mu cos30° = -0.4v cos45° | A1 | |
| a = 0.318 | A1 [5] |
| Answer | Marks | Guidance |
|---|---|---|
| [e(3cosα + u cos30°) = v cos45°] | M1 | For using N.E.L. |
| 0.75(4cos\(\theta\) + u cos30°) = v cos45° | B1 | |
| 4sin\(\theta\) = v sin45° | M1 | |
| [3cos\(\theta\) + 4.5cos30° = 4sin\(\theta\)] | A1 | AG |
| 8sin\(\theta\) - 6cos\(\theta\) = 9cos30° | A1 [5] |
| Answer | Marks | Guidance |
|---|---|---|
| Extension = 1.2α - 0.6 [T = mg\sinα] | B1 M1 | For resolving forces tangentially |
| 0.5x9.8\sin α = 6.86(1.2α - 0.6)/0.6 | A1ft AG |
| Answer | Marks | Guidance |
|---|---|---|
| [0.8, 0.756..., 0.745..., 0.742...] [0.741..., 0.741...] | M1 | For attempting to find α₂ and α₃ |
| α = 0.74 | A1 [2] |
**(i)**
| [u sin30° = 3] | M1 | For momentum equation for B, normal to line of centres |
|---|---|---|
| u = 6 | A1 [2] |
**(ii)**
| [4sin88.1° = v sin45°] | M1 | For momentum equation for A, normal to line of centres |
|---|---|---|
| v = 5.65 | A1 M1 | For momentum equation along line of centres |
|---|---|---|
| 0.4(4cos88.1°) - mu cos30° = -0.4v cos45° | A1 | |
|---|---|---|
| a = 0.318 | A1 [5] |
**(iii)**
| [e(3cosα + u cos30°) = v cos45°] | M1 | For using N.E.L. |
|---|---|---|
| 0.75(4cos$\theta$ + u cos30°) = v cos45° | B1 | |
|---|---|---|
| 4sin$\theta$ = v sin45° | M1 | |
|---|---|---|
| [3cos$\theta$ + 4.5cos30° = 4sin$\theta$] | A1 | AG |
|---|---|---|
| 8sin$\theta$ - 6cos$\theta$ = 9cos30° | A1 [5] |
**(iv)**
| Extension = 1.2α - 0.6 [T = mg\sinα] | B1 M1 | For resolving forces tangentially |
|---|---|---|
| 0.5x9.8\sin α = 6.86(1.2α - 0.6)/0.6 | A1ft AG | |
**(iv)(b)**
| [0.8, 0.756..., 0.745..., 0.742...] [0.741..., 0.741...] | M1 | For attempting to find α₂ and α₃ |
|---|---|---|
| α = 0.74 | A1 [2] |6\\
\includegraphics[max width=\textwidth, alt={}, center]{af1f9f1b-f6c0-4044-9864-5b9ce309d3fa-03_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 Q6 [14]}}