| Exam Board | AQA |
|---|---|
| Module | M2 (Mechanics 2) |
| Year | 2015 |
| Session | June |
| Marks | 10 |
| 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 M2 circular motion problem with two strings. Part (a) requires resolving vertically (straightforward equilibrium), part (b) involves resolving horizontally and algebraic manipulation to reach a given answer, and part (c) solves simultaneous equations. While it requires careful resolution of forces and understanding of circular motion, it follows a well-established method with no novel insight needed. Slightly easier than average due to the structured parts and 'show that' guidance. |
| Spec | 3.03n Equilibrium in 2D: particle under forces6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Resolving vertically: \(T_{AP}\cos 20° = 5g\) | M1 | Resolving vertically with \(T_{AP}\) |
| \(T_{AP} = \frac{5g}{\cos 20°}\) | A1 | Correct equation |
| \(T_{AP} = 52.3\) N | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Horizontal equation: \(T_{AP}\sin 20° - T_{BP} = \frac{mv^2}{r} = \frac{5v^2}{0.6}\) | M1 | Applying Newton's second law horizontally |
| \(T_{BP} = T_{AP}\sin 20° - \frac{5v^2}{0.6}\) | A1 | Correct equation |
| \(T_{BP} = \frac{5g}{\cos 20°}\sin 20° - \frac{25v^2}{3} = 5g\tan 20° - \frac{25}{3}v^2\) | A1 | Correctly shown (given result) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Setting \(T_{AP} = T_{BP}\): \(\frac{5g}{\cos 20°} = \frac{25}{3}v^2 - 5g\tan 20°\) | M1 | Equating the two tensions |
| \(\frac{5g}{\cos 20°} + 5g\tan 20° = \frac{25}{3}v^2\) | M1 | Rearranging |
| \(v^2 = \frac{3}{25}\left(\frac{5g}{\cos 20°} + 5g\tan 20°\right)\) | A1 | Correct expression |
| \(v = 2.74\) m s\(^{-1}\) | A1 | cao |
# Question 4:
## Part (a)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Resolving vertically: $T_{AP}\cos 20° = 5g$ | M1 | Resolving vertically with $T_{AP}$ |
| $T_{AP} = \frac{5g}{\cos 20°}$ | A1 | Correct equation |
| $T_{AP} = 52.3$ N | A1 | cao |
## Part (b)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Horizontal equation: $T_{AP}\sin 20° - T_{BP} = \frac{mv^2}{r} = \frac{5v^2}{0.6}$ | M1 | Applying Newton's second law horizontally |
| $T_{BP} = T_{AP}\sin 20° - \frac{5v^2}{0.6}$ | A1 | Correct equation |
| $T_{BP} = \frac{5g}{\cos 20°}\sin 20° - \frac{25v^2}{3} = 5g\tan 20° - \frac{25}{3}v^2$ | A1 | Correctly shown (given result) |
## Part (c)
| Answer/Working | Marks | Guidance |
|---|---|---|
| Setting $T_{AP} = T_{BP}$: $\frac{5g}{\cos 20°} = \frac{25}{3}v^2 - 5g\tan 20°$ | M1 | Equating the two tensions |
| $\frac{5g}{\cos 20°} + 5g\tan 20° = \frac{25}{3}v^2$ | M1 | Rearranging |
| $v^2 = \frac{3}{25}\left(\frac{5g}{\cos 20°} + 5g\tan 20°\right)$ | A1 | Correct expression |
| $v = 2.74$ m s$^{-1}$ | A1 | cao |
---4 A particle, $P$, of mass 5 kg is attached to two light inextensible strings, $A P$ and $B P$. The other ends of the strings are attached to the fixed points $A$ and $B$. The point $A$ is vertically above the point $B$. The particle moves at a constant speed, $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$, in a horizontal circle of radius 0.6 metres with centre $B$. The string $A P$ is inclined at $20 ^ { \circ }$ to the vertical, as shown in the diagram. Both strings are taut when the particle is moving.\\
\includegraphics[max width=\textwidth, alt={}, center]{691c50b4-50b2-4e3a-a7e0-60f8ec35ee3c-08_835_568_568_719}
\begin{enumerate}[label=(\alph*)]
\item Find the tension in the string $A P$.
\item The speed of the particle is $v \mathrm {~m} \mathrm {~s} ^ { - 1 }$.
Show that the tension, $T _ { B P }$, in the string $B P$ is given by
$$T _ { B P } = \frac { 25 } { 3 } v ^ { 2 } - 5 g \tan 20 ^ { \circ }$$
\item Find $v$ when the tensions in the two strings are equal.
\end{enumerate}
\hfill \mbox{\textit{AQA M2 2015 Q4 [10]}}