| Exam Board | Edexcel |
|---|---|
| Module | FM2 (Further Mechanics 2) |
| Year | 2021 |
| Session | June |
| Marks | 10 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Smooth ring on rotating string |
| Difficulty | Standard +0.8 This is a Further Mechanics 2 circular motion problem requiring 3D geometric visualization, simultaneous resolution of forces in two directions, and application of circular motion principles. While the setup is standard for FM2, students must correctly identify the geometry (Pythagorean theorem to find BR), resolve tensions in both vertical and horizontal components, and apply F=mrω². The multi-step nature and spatial reasoning elevate it above average A-level difficulty, though it follows a recognizable FM2 pattern. |
| Spec | 3.03d Newton's second law: 2D vectors3.03f Weight: W=mg6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(l^2 + r^2 = (2l-r)^2\), using Pythagoras | M1 | Use of Pythagoras with one unknown |
| \(BR = \frac{3l}{4}\) | A1* | Correct length |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Resolve vertically | M1 | Allow sin/cos confusion |
| \(T\cos\alpha = mg\) | A1 | Correct equation |
| Substitute for \(\cos\alpha\) and solve for \(T\) | M1 | Substituting for their trig ratio and solving for \(T\) |
| \(T = \frac{5mg}{4}\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Equation of motion horizontally | M1 | Correct no. of terms, dimensionally correct |
| \(T + T\sin\alpha = \frac{mV^2}{r}\) | A1 | Correct equation |
| Substitute for \(T\), \(\sin\alpha\) and \(r\) and solve for \(V\) | M1 | Substitute for \(T\), \(\sin\alpha\) and \(r\) and solve for \(V\) |
| \(V = \sqrt{\frac{3gl}{2}}\) | A1 | cao. Accept other equivalent forms |
## Question 4(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $l^2 + r^2 = (2l-r)^2$, using Pythagoras | M1 | Use of Pythagoras with one unknown |
| $BR = \frac{3l}{4}$ | A1* | Correct length |
## Question 4(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Resolve vertically | M1 | Allow sin/cos confusion |
| $T\cos\alpha = mg$ | A1 | Correct equation |
| Substitute for $\cos\alpha$ and solve for $T$ | M1 | Substituting for their trig ratio and solving for $T$ |
| $T = \frac{5mg}{4}$ | A1 | cao |
## Question 4(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Equation of motion horizontally | M1 | Correct no. of terms, dimensionally correct |
| $T + T\sin\alpha = \frac{mV^2}{r}$ | A1 | Correct equation |
| Substitute for $T$, $\sin\alpha$ and $r$ and solve for $V$ | M1 | Substitute for $T$, $\sin\alpha$ and $r$ and solve for $V$ |
| $V = \sqrt{\frac{3gl}{2}}$ | A1 | cao. Accept other equivalent forms |4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d4fc2ea6-3ffc-42f2-b462-9694adfe2ec1-14_682_817_246_625}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
One end of a light inextensible string of length $2 l$ is attached to a fixed point $A$. A small smooth ring $R$ of mass $m$ is threaded on the string and the other end of the string is attached to a fixed point $B$. The point $B$ is vertically below $A$, with $A B = l$. The ring is then made to move with constant speed $V$ in a horizontal circle with centre $B$. The string is taut and $B R$ is horizontal, as shown in Figure 4.
\begin{enumerate}[label=(\alph*)]
\item Show that $B R = \frac { 31 } { 4 }$
Given that air resistance is negligible,
\item find, in terms of $m$ and $g$, the tension in the string,
\item find $V$ in terms of $g$ and $l$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FM2 2021 Q4 [10]}}