| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2006 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Particle on table with string above |
| Difficulty | Standard +0.3 This is a standard M3 circular motion problem with clear geometry and straightforward application of Newton's second law in both horizontal and vertical directions. Part (a) is a 'show that' requiring resolving forces, part (b) follows directly, part (c) requires recognizing R≥0, and part (d) involves recalculating with new conditions. All steps are routine for M3 students with no novel insights required. |
| Spec | 3.03n Equilibrium in 2D: particle under forces6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| N2L \(\leftarrow\): \(T\cos 30° = m(2a\cos 30°)\left(\dfrac{kg}{3a}\right)\) | M1 A1 | |
| \(T = \dfrac{2kmg}{3}\) ✱ | A1 | cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\uparrow\; R = mg - T\sin 30°\) | M1 A1 | |
| \(= mg\left(1 - \dfrac{k}{3}\right)\) | A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \((R \geq 0) \Rightarrow k \leq 3\) | M1 A1 | ignore \(k > 0\), accept \(k < 3\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| N2L \(\leftarrow\): \(T\cos\theta = m(2a\cos\theta)\left(\dfrac{2g}{a}\right)\), \((T = 4mg)\) | M1 A1 | |
| \(\uparrow\; T\sin\theta = mg\) | M1 | |
| Eliminating \(T\) | M1 | |
| \(AX = 2a\sin\theta = \dfrac{1}{2}a\) | A1 | |
| \(AO = 2a\sin 30° = a \Rightarrow AX = \dfrac{1}{2}AO\) ✱ | B1, A1 | cso |
## Question 7:
**Part (a)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| N2L $\leftarrow$: $T\cos 30° = m(2a\cos 30°)\left(\dfrac{kg}{3a}\right)$ | M1 A1 | |
| $T = \dfrac{2kmg}{3}$ ✱ | A1 | cso |
**(3 marks)**
**Part (b)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\uparrow\; R = mg - T\sin 30°$ | M1 A1 | |
| $= mg\left(1 - \dfrac{k}{3}\right)$ | A1 | |
**(3 marks)**
**Part (c)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $(R \geq 0) \Rightarrow k \leq 3$ | M1 A1 | ignore $k > 0$, accept $k < 3$ |
**(2 marks)**
**Part (d)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| N2L $\leftarrow$: $T\cos\theta = m(2a\cos\theta)\left(\dfrac{2g}{a}\right)$, $(T = 4mg)$ | M1 A1 | |
| $\uparrow\; T\sin\theta = mg$ | M1 | |
| Eliminating $T$ | M1 | |
| $AX = 2a\sin\theta = \dfrac{1}{2}a$ | A1 | |
| $AO = 2a\sin 30° = a \Rightarrow AX = \dfrac{1}{2}AO$ ✱ | B1, A1 | cso |
**(7 marks) — Total 15 marks**
7.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 3}
\includegraphics[alt={},max width=\textwidth]{67a9cf74-833f-4b4a-9fde-3c62dcc08e8c-5_625_1141_319_424}
\end{center}
\end{figure}
A particle $P$ of mass $m$ is attached to one end of a light inextensible string of length $2 a$. The other end of the string is fixed to a point $A$ which is vertically above the point $O$ on a smooth horizontal table. The particle $P$ remains in contact with the surface of the table and moves in a circle with centre $O$ and with angular speed $\sqrt { \left( \frac { k g } { 3 a } \right) }$, where $k$ is a constant. Throughout the motion the string remains taut and $\angle A P O = 30 ^ { \circ }$, as shown in Figure 3.
\begin{enumerate}[label=(\alph*)]
\item Show that the tension in the string is $\frac { 2 k m g } { 3 }$.
\item Find, in terms of $m , g$ and $k$, the normal reaction between $P$ and the table.
\item Deduce the range of possible values of $k$.
The angular speed of $P$ is changed to $\sqrt { \left( \frac { 2 g } { a } \right) }$. The particle $P$ now moves in a horizontal circle above the table. The centre of this circle is $X$.
\item Show that $X$ is the mid-point of $O A$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2006 Q7 [15]}}