| Exam Board | Edexcel |
|---|---|
| Module | FM2 AS (Further Mechanics 2 AS) |
| Year | 2021 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Particle on table with string above |
| Difficulty | Challenging +1.2 This is a standard Further Mechanics circular motion problem requiring resolution of forces in a conical pendulum setup. Part (a) involves routine application of Newton's second law with given angular speed, part (b) requires finding the critical condition when normal reaction becomes zero, and part (c) tests conceptual understanding. The geometry is straightforward (3-4-5 triangle) and the problem follows a familiar textbook pattern, though it requires more steps than a typical A-level question and involves Further Maths content. |
| Spec | 6.05b Circular motion: v=r*omega and a=v^2/r6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Resolve vertically | M1 | Forces balance vertically. All terms required. Condone sign errors and sin/cos confusion |
| \(T\sin\theta + N = mg\) | A1 | Correct unsimplified equation |
| Resolve horizontally | M1 | Equation for motion in a horizontal circle. All terms required. Condone sign errors and sin/cos confusion |
| \(T + T\cos\theta = ma \times \dfrac{2g}{3a}\) giving \(\dfrac{4}{3}T = \dfrac{2}{3}mg\), \(T = \dfrac{1}{2}mg\) | A1 | Correct unsimplified equation |
| Complete strategy to obtain and solve equations to find \(N\) | M1 | Complete strategy to use the model to form simultaneous equations and solve for \(N\) |
| \(\dfrac{\sqrt{8}}{3} \times \dfrac{1}{2}mg + N = mg\), \(N = mg\left(1 - \dfrac{\sqrt{2}}{3}\right)\) | A1 | Any equivalent form. Accept \(0.53mg\) or better |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Max speed \(\Rightarrow N = 0\), \(\dfrac{\sqrt{8}}{3}T = mg\) | M1 | Resolve vertically and use max speed \(\Rightarrow N = 0\) to form equation in \(T\) |
| \(\dfrac{4}{3}T = ma\omega^2 \Rightarrow \dfrac{3}{\sqrt{8}}mg \times \dfrac{4}{3} = ma\omega^2\) | M1 | Resolve horizontally and substitute for \(T\) to form equation in \(\omega\) |
| \(\omega = \sqrt{\dfrac{4g}{\sqrt{8}a}} = \sqrt{\dfrac{\sqrt{2}g}{a}}\) | A1 | Any equivalent simplified form. \(1.2\sqrt{\dfrac{g}{a}}\) or better |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| The tension in the string is the same on either side of \(P\) | B1 | Clear statement explaining how the modelling assumption has been used |
## Question 2:
**Part (a):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolve vertically | M1 | Forces balance vertically. All terms required. Condone sign errors and sin/cos confusion |
| $T\sin\theta + N = mg$ | A1 | Correct unsimplified equation |
| Resolve horizontally | M1 | Equation for motion in a horizontal circle. All terms required. Condone sign errors and sin/cos confusion |
| $T + T\cos\theta = ma \times \dfrac{2g}{3a}$ giving $\dfrac{4}{3}T = \dfrac{2}{3}mg$, $T = \dfrac{1}{2}mg$ | A1 | Correct unsimplified equation |
| Complete strategy to obtain and solve equations to find $N$ | M1 | Complete strategy to use the model to form simultaneous equations and solve for $N$ |
| $\dfrac{\sqrt{8}}{3} \times \dfrac{1}{2}mg + N = mg$, $N = mg\left(1 - \dfrac{\sqrt{2}}{3}\right)$ | A1 | Any equivalent form. Accept $0.53mg$ or better |
**Part (b):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| Max speed $\Rightarrow N = 0$, $\dfrac{\sqrt{8}}{3}T = mg$ | M1 | Resolve vertically and use max speed $\Rightarrow N = 0$ to form equation in $T$ |
| $\dfrac{4}{3}T = ma\omega^2 \Rightarrow \dfrac{3}{\sqrt{8}}mg \times \dfrac{4}{3} = ma\omega^2$ | M1 | Resolve horizontally and substitute for $T$ to form equation in $\omega$ |
| $\omega = \sqrt{\dfrac{4g}{\sqrt{8}a}} = \sqrt{\dfrac{\sqrt{2}g}{a}}$ | A1 | Any equivalent simplified form. $1.2\sqrt{\dfrac{g}{a}}$ or better |
**Part (c):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| The tension in the string is the same on either side of $P$ | B1 | Clear statement explaining how the modelling assumption has been used |2.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{a7901165-1679-4d30-9444-0c27020e32ea-04_572_889_246_589}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A small smooth ring $P$, of mass $m$, is threaded onto a light inextensible string of length 4a. One end of the string is attached to a fixed point $A$ on a smooth horizontal table. The other end of the string is attached to a fixed point $B$ which is vertically above $A$. The ring moves in a horizontal circle with centre $A$ and radius $a$, as shown in Figure 2.
The ring moves with constant angular speed $\sqrt { \frac { 2 g } { 3 a } }$ about $A B$.\\
The string remains taut throughout the motion.
\begin{enumerate}[label=(\alph*)]
\item Find, in terms of $m$ and $g$, the magnitude of the normal reaction between $P$ and the table.
The angular speed of $P$ is now gradually increased.
\item Find, in terms of $a$ and $g$, the angular speed of $P$ at the instant when it loses contact with the table.
\item Explain how you have used the fact that $P$ is smooth.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FM2 AS 2021 Q2 [10]}}