| Exam Board | Edexcel |
|---|---|
| Module | FM2 (Further Mechanics 2) |
| Year | 2020 |
| Session | June |
| Marks | 11 |
| 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 conical pendulum problem with a table constraint. Part (a) requires routine application of circular motion equations (F=mrω²) and vertical equilibrium, involving basic trigonometry with a 3-4-5 triangle. Part (b) requires recognizing that contact is lost when normal reaction becomes zero, leading to a straightforward calculation. While it's a multi-part question requiring several steps, all techniques are standard for FM2 circular motion with no novel insight needed, making it slightly easier than average. |
| Spec | 3.03d Newton's second law: 2D vectors3.03f Weight: W=mg3.03i Normal reaction force6.05c Horizontal circles: conical pendulum, banked tracks |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Resolve vertically | M1 | 3.4 - Correct number of terms |
| \(0.75g = T\cos\theta + R\), \(\left(\frac{3g}{4} = \frac{2}{3}T + R\right)\) | A1 | 1.1b - Correct unsimplified equation |
| Equation of motion: \(0.75 \times \sin\theta \times 9 = T\sin\theta\), \(\left(0.75 \times \frac{\sqrt{20}}{10} \times 9 = T \times \frac{\sqrt{20}}{6}\right)\) | M1 | 3.4 - Circular motion; condone confusion over units; \(\frac{\sqrt{20}}{10}\) might not be seen as \(r\) cancels |
| Correct unsimplified equation | A1 | 1.1b - Correct unsimplified equation |
| Complete strategy to find \(T\) and \(R\) | M1 | 3.1a - Complete strategy to form sufficient equations to solve for \(T\) and \(R\) |
| \(T = \frac{6 \times 0.75 \times 9}{10} = 4.05\) (N) | A1 | 1.1b - One force correct |
| \(R = 0.75g - \frac{2}{3}T = 4.65\) (N) or \(4.7\) (N) | A1 | 1.1b - Both correct (finding value for \(R\) involves \(g\)) |
| Total: (7) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Use \(R = 0\) to form revised equations | M1 | 3.4 - Correct interpretation of loss of contact |
| \(T\cos\theta = 0.75g\), \(T\sin\theta = 0.75 \times \frac{10\sqrt{20}}{100}\omega^2\) (or \(T\sin\theta = 0.75 \times 0.6\sin\theta \times \omega^2\)) | A1 | 1.1b - Revised equations |
| Complete strategy to find \(\omega\), e.g. \(\Rightarrow \tan\theta = \frac{\sqrt{20}\omega^2}{10g} = \frac{\sqrt{20}}{4}\) | M1 | 1.1b - Solve for \(\omega\) |
| \(\omega = \sqrt{\frac{5g}{2}} = 4.95\) (rad/s) | A1 | 1.1b - Exact, 4.9 or 4.95 (non-exact answer requires substitution for \(g\)) |
| Total: (4) |
# Question 5(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Resolve vertically | M1 | 3.4 - Correct number of terms |
| $0.75g = T\cos\theta + R$, $\left(\frac{3g}{4} = \frac{2}{3}T + R\right)$ | A1 | 1.1b - Correct unsimplified equation |
| Equation of motion: $0.75 \times \sin\theta \times 9 = T\sin\theta$, $\left(0.75 \times \frac{\sqrt{20}}{10} \times 9 = T \times \frac{\sqrt{20}}{6}\right)$ | M1 | 3.4 - Circular motion; condone confusion over units; $\frac{\sqrt{20}}{10}$ might not be seen as $r$ cancels |
| Correct unsimplified equation | A1 | 1.1b - Correct unsimplified equation |
| Complete strategy to find $T$ and $R$ | M1 | 3.1a - Complete strategy to form sufficient equations to solve for $T$ and $R$ |
| $T = \frac{6 \times 0.75 \times 9}{10} = 4.05$ (N) | A1 | 1.1b - One force correct |
| $R = 0.75g - \frac{2}{3}T = 4.65$ (N) or $4.7$ (N) | A1 | 1.1b - Both correct (finding value for $R$ involves $g$) |
| **Total: (7)** | | |
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# Question 5(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Use $R = 0$ to form revised equations | M1 | 3.4 - Correct interpretation of loss of contact |
| $T\cos\theta = 0.75g$, $T\sin\theta = 0.75 \times \frac{10\sqrt{20}}{100}\omega^2$ (or $T\sin\theta = 0.75 \times 0.6\sin\theta \times \omega^2$) | A1 | 1.1b - Revised equations |
| Complete strategy to find $\omega$, e.g. $\Rightarrow \tan\theta = \frac{\sqrt{20}\omega^2}{10g} = \frac{\sqrt{20}}{4}$ | M1 | 1.1b - Solve for $\omega$ |
| $\omega = \sqrt{\frac{5g}{2}} = 4.95$ (rad/s) | A1 | 1.1b - Exact, 4.9 or 4.95 (non-exact answer requires substitution for $g$) |
| **Total: (4)** | | |
---5.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{962c2b40-3c45-4eed-a0af-a59068bda0e1-16_501_606_244_731}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
A particle $P$ of mass 0.75 kg is attached to one end of a light inextensible string of length 60 cm . The other end of the string is attached to a fixed point $A$ that is vertically above the point $O$ on a smooth horizontal table, such that $O A = 40 \mathrm {~cm}$. The particle remains in contact with the table, with the string taut, and moves in a horizontal circle with centre $O$, as shown in Figure 4.
The particle is moving with a constant angular speed of 3 radians per second.
\begin{enumerate}[label=(\alph*)]
\item Find (i) the tension in the string,\\
(ii) the normal reaction between $P$ and the table.
The angular speed of $P$ is now gradually increased.
\item Find the angular speed of $P$ at the instant $P$ loses contact with the table.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FM2 2020 Q5 [11]}}