| Exam Board | OCR |
|---|---|
| Module | M2 (Mechanics 2) |
| Session | Specimen |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Two strings, two fixed points |
| Difficulty | Standard +0.3 This is a standard circular motion problem requiring resolution of forces and application of F=ma=mv²/r. Part (i) involves straightforward simultaneous equations with vertical equilibrium and horizontal centripetal force. Part (ii) requires recognizing that the lower string becomes slack when its tension reaches zero, which is a common textbook scenario. The geometry is given explicitly, making this slightly easier than average. |
| Spec | 3.03n Equilibrium in 2D: particle under forces6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| (i) \(T_1 \times \frac{2}{25} = 0.08g\). Hence tension in upper string is 2.8 N. \(T_1 \times \frac{24}{25} + T_2 = 0.08 \times \frac{10.5^2}{2.4}\) | M1, B1, A1, M1, B1, A1 (7 marks) | For resolving vertically; For \(\frac{2}{25}\) or \(\sin16.3°\) or equivalent; For correct value 2.8; For correct use of Newton II horizontally; For any use of \(\frac{10.5^2}{2.4}\), or equivalent; For correct horizontal equation |
| Hence tension in horizontal string is 0.987 N | A1 (1 mark) | For correct value 0.987 |
| (ii) \(2.8 \times \frac{2.4}{2.5} = 0.08 \times \frac{v^2}{2.4}\). Hence \(v = 8.98\) | M1, A1, M1, A1 (4 marks) | For new horizontal equation with \(T_2 = 0\); For correct equation for \(v\); For solving for \(v\) correctly; For correct value 8.98 |
(i) $T_1 \times \frac{2}{25} = 0.08g$. Hence tension in upper string is 2.8 N. $T_1 \times \frac{24}{25} + T_2 = 0.08 \times \frac{10.5^2}{2.4}$ | M1, B1, A1, M1, B1, A1 (7 marks) | For resolving vertically; For $\frac{2}{25}$ or $\sin16.3°$ or equivalent; For correct value 2.8; For correct use of Newton II horizontally; For any use of $\frac{10.5^2}{2.4}$, or equivalent; For correct horizontal equation
Hence tension in horizontal string is 0.987 N | A1 (1 mark) | For correct value 0.987
(ii) $2.8 \times \frac{2.4}{2.5} = 0.08 \times \frac{v^2}{2.4}$. Hence $v = 8.98$ | M1, A1, M1, A1 (4 marks) | For new horizontal equation with $T_2 = 0$; For correct equation for $v$; For solving for $v$ correctly; For correct value 8.98
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\includegraphics[max width=\textwidth, alt={}, center]{b96a99a6-3df4-4000-9bf1-aab7ab954b4a-4_314_757_285_708}
A ball of mass 0.08 kg is attached by two strings to a fixed vertical post. The strings have lengths 2.5 m and 2.4 m , as shown in the diagram. The ball moves in a horizontal circle, of radius 2.4 m , with constant speed $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$. Each string is taut and the lower string is horizontal. The modelling assumptions made are that both strings are light and inextensible, and that there is no air resistance.\\
(i) Find the tension in each string when $v = 10.5$.\\
(ii) Find the least value of $v$ for which the lower string is taut.
\hfill \mbox{\textit{OCR M2 Q7 [11]}}