| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2013 |
| Session | January |
| Marks | 11 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Elastic string – conical pendulum (string inclined to vertical) |
| Difficulty | Standard +0.3 This is a standard M3 circular motion problem with elastic string requiring resolution of forces and Hooke's law. The geometry is straightforward (3-4-5 triangle), and both parts follow a routine method: (a) uses vertical equilibrium to find extension, (b) uses horizontal equation for circular motion. Slightly easier than average due to the clear setup and standard technique. |
| Spec | 6.02h Elastic PE: 1/2 k x^26.02i Conservation of energy: mechanical energy principle6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(R(\uparrow)\quad T\cos\theta = mg\) | M1 | |
| \(T\times\frac{2a}{(2a+x)} = mg\) | A1 | |
| Hooke's Law: \(T = \frac{6mgx}{2a} = \frac{3mgx}{a}\) | M1A1 | |
| \(\frac{3mgx}{a}\times\frac{2a}{(2a+x)} = mg\) | M1dep | |
| \(6x = 2a+x\) | ||
| \(x = \frac{2}{5}a\) | A1 | Given/shown result |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Notes |
| \(T\sin\theta = \frac{mv^2}{r}\) | M1A1 | |
| \(3mg\times\frac{2}{5}\sin\theta = \frac{mv^2}{\left(\frac{12a}{5}\right)\sin\theta}\) | M1dep | |
| \(v^2 = \frac{6}{5}g\times\frac{12a}{5}\sin^2\theta\) | ||
| \(\sin^2\theta = 1-\left(\frac{4a^2}{\left(\frac{12a}{5}\right)^2}\right) = \frac{11}{36}\) | ||
| \(v^2 = \frac{72ag}{25}\times\frac{11}{36} = \frac{22ag}{25}\) | M1depA1 |
## Question 4:
### Part (a):
| Answer/Working | Marks | Notes |
|---|---|---|
| $R(\uparrow)\quad T\cos\theta = mg$ | M1 | |
| $T\times\frac{2a}{(2a+x)} = mg$ | A1 | |
| Hooke's Law: $T = \frac{6mgx}{2a} = \frac{3mgx}{a}$ | M1A1 | |
| $\frac{3mgx}{a}\times\frac{2a}{(2a+x)} = mg$ | M1dep | |
| $6x = 2a+x$ | | |
| $x = \frac{2}{5}a$ | A1 | Given/shown result |
### Part (b):
| Answer/Working | Marks | Notes |
|---|---|---|
| $T\sin\theta = \frac{mv^2}{r}$ | M1A1 | |
| $3mg\times\frac{2}{5}\sin\theta = \frac{mv^2}{\left(\frac{12a}{5}\right)\sin\theta}$ | M1dep | |
| $v^2 = \frac{6}{5}g\times\frac{12a}{5}\sin^2\theta$ | | |
| $\sin^2\theta = 1-\left(\frac{4a^2}{\left(\frac{12a}{5}\right)^2}\right) = \frac{11}{36}$ | | |
| $v^2 = \frac{72ag}{25}\times\frac{11}{36} = \frac{22ag}{25}$ | M1depA1 | |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d19c7390-0332-4cab-82e5-72976bd499a2-07_503_618_242_646}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A particle $P$ of mass $m$ is attached to one end of a light elastic string, of natural length $2 a$ and modulus of elasticity $6 m g$. The other end of the string is attached to a fixed point $A$. The particle moves with constant speed $v$ in a horizontal circle with centre $O$, where $O$ is vertically below $A$ and $O A = 2 a$, as shown in Figure 2 .
\begin{enumerate}[label=(\alph*)]
\item Show that the extension in the string is $\frac { 2 } { 5 } a$.
\item Find $v ^ { 2 }$ in terms of $a$ and $g$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2013 Q4 [11]}}