| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2007 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | String through hole – hanging particle in equilibrium below table |
| Difficulty | Standard +0.3 This is a standard M2 circular motion problem with coupled particles. Part (a) requires recognizing equilibrium of Q (T = 5g = 49N), part (b) uses horizontal circular motion equation (T cos θ = mv²/r and T sin θ = 3g) to find θ, and part (c) finds radius from geometry. All steps are routine applications of standard mechanics principles with no novel insight required, making it slightly easier than average. |
| Spec | 6.03c Momentum in 2D: vector form6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Part (a): \(Q\) is in equilibrium; \(T = 5g = 49\) N | E1, B1 | Q at rest, or not moving; AG |
| Part (b): Resolving vertically for \(P\): \(T\cos\theta = 3g\); \(\cos\theta = \frac{3}{5}\); \(\theta = \cos^{-1}\frac{3}{5} = 53.1°\) | M1A1, A1 | Do not condone 53° |
| Part (c): \(\therefore \sin\theta = \frac{4}{5}\); Resolving horizontally for \(P\): \(\frac{mv^2}{r} = T\sin\theta\); \(\frac{3v^2}{r} = \frac{4}{5} \times 5g\); \(\frac{3 \times 4^2}{r} = 4g\); \(r = \frac{48}{4g} = 1.22\) | B1, M1A1, A1 | M1 2 terms: 1 term correct, other term includes sin or cos; SC3 1·23 |
| Total: 9 |
| Answer/Working | Marks | Guidance |
|---|---|---|
| **Part (a):** $Q$ is in equilibrium; $T = 5g = 49$ N | E1, B1 | Q at rest, or not moving; AG |
| **Part (b):** Resolving vertically for $P$: $T\cos\theta = 3g$; $\cos\theta = \frac{3}{5}$; $\theta = \cos^{-1}\frac{3}{5} = 53.1°$ | M1A1, A1 | Do not condone 53° |
| **Part (c):** $\therefore \sin\theta = \frac{4}{5}$; Resolving horizontally for $P$: $\frac{mv^2}{r} = T\sin\theta$; $\frac{3v^2}{r} = \frac{4}{5} \times 5g$; $\frac{3 \times 4^2}{r} = 4g$; $r = \frac{48}{4g} = 1.22$ | B1, M1A1, A1 | M1 2 terms: 1 term correct, other term includes sin or cos; SC3 1·23 |
| | | **Total: 9** |
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## TOTAL: 758 A particle, $P$, of mass 3 kg is attached to one end of a light inextensible string. The string passes through a smooth fixed ring, $O$, and a second particle, $Q$, of mass 5 kg is attached to the other end of the string. The particle $Q$ hangs at rest vertically below the ring and the particle $P$ moves with speed $4 \mathrm {~m} \mathrm {~s} ^ { - 1 }$ in a horizontal circle, as shown in the diagram.
The angle between $O P$ and the vertical is $\theta$.\\
\includegraphics[max width=\textwidth, alt={}, center]{676e753d-1b80-413c-a4b9-21861db8dde5-5_474_476_1425_774}
\begin{enumerate}[label=(\alph*)]
\item Explain why the tension in the string is 49 N .
\item Find $\theta$.
\item Find the radius of the horizontal circle.
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2007 Q8 [9]}}