| Exam Board | Edexcel |
|---|---|
| Module | FM2 (Further Mechanics 2) |
| Session | Specimen |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Vertical circle – string/rod (tension and energy) |
| Difficulty | Standard +0.8 This is a standard Further Maths vertical circular motion problem requiring energy conservation, circular motion dynamics, and understanding of contact forces. While it involves multiple parts and careful application of principles, the techniques are well-practiced in FM2 and follow predictable patterns. The 'show that' in part (a) guides students, and parts (b)-(d) are direct applications of standard methods, though part (d) requires careful vector consideration of acceleration components. |
| Spec | 3.02h Motion under gravity: vector form6.02i Conservation of energy: mechanical energy principle6.05d Variable speed circles: energy methods6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Conservation of energy | M1 | All terms required. Must be dimensionally correct |
| \(\frac{1}{2}mv^2 + mga(1-\cos\theta) = \frac{1}{2}m\left(\frac{7}{2}ga\right)\) | A1 | Correct unsimplified equation |
| \(v^2 = ga\left(\frac{3}{2} + 2\cos\theta\right)\)* | A1* | Show sufficient working to justify given answer and a 'statement' that the required result has been achieved |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Resolve parallel to \(OB\) and use \(\frac{mv^2}{a}\) | M1 | Resolve parallel to \(OB\) |
| \(R - mg\cos\theta = \frac{mv^2}{a}\) | A1 | Correct equation |
| Use \(R=0 \Rightarrow g\cos\theta = -\frac{v^2}{a}\) | M1 | Use \(R=0\) seen or implied |
| Solve for \(\theta \Rightarrow g\cos\theta = -g\left(\frac{3}{2} + 2\cos\theta\right)\) | M1 | Solve for \(\theta\) |
| \(\theta = 120°\) | A1 | Accept \(\frac{2\pi}{3}\) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Any appropriate comment e.g. the hoop is unlikely to be smooth | B1 | Any appropriate comment e.g. hoop may not be smooth; air resistance could affect the motion |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| At rest \(\Rightarrow v = 0\) | M1 | \(v=0\) seen or implied |
| \(\Rightarrow \cos\theta = -\frac{3}{4}\) | A1 | Correct equation in \(\theta\) |
| Acceleration is tangential | M1 | Correct direction for acceleration |
| Magnitude \(\left | g\cos(\theta - 90)\right | = 6.48\ \text{m s}^{-2}\) or \(\frac{\sqrt{7}}{4}g\) |
| At \(\left(\cos^{-1}\left(-\frac{3}{4}\right) - 90 =\right) 48.6°\) to the downward vertical | A1 | Accept 0.848 (radians) |
# Question 6:
## Part (a)
| Working/Answer | Mark | Guidance |
|---|---|---|
| Conservation of energy | M1 | All terms required. Must be dimensionally correct |
| $\frac{1}{2}mv^2 + mga(1-\cos\theta) = \frac{1}{2}m\left(\frac{7}{2}ga\right)$ | A1 | Correct unsimplified equation |
| $v^2 = ga\left(\frac{3}{2} + 2\cos\theta\right)$* | A1* | Show sufficient working to justify given answer and a 'statement' that the required result has been achieved |
**(3 marks)**
## Part (b)
| Working/Answer | Mark | Guidance |
|---|---|---|
| Resolve parallel to $OB$ and use $\frac{mv^2}{a}$ | M1 | Resolve parallel to $OB$ |
| $R - mg\cos\theta = \frac{mv^2}{a}$ | A1 | Correct equation |
| Use $R=0 \Rightarrow g\cos\theta = -\frac{v^2}{a}$ | M1 | Use $R=0$ seen or implied |
| Solve for $\theta \Rightarrow g\cos\theta = -g\left(\frac{3}{2} + 2\cos\theta\right)$ | M1 | Solve for $\theta$ |
| $\theta = 120°$ | A1 | Accept $\frac{2\pi}{3}$ |
**(5 marks)**
## Part (c)
| Working/Answer | Mark | Guidance |
|---|---|---|
| Any appropriate comment e.g. the hoop is unlikely to be smooth | B1 | Any appropriate comment e.g. hoop may not be smooth; air resistance could affect the motion |
**(1 mark)**
## Part (d)
| Working/Answer | Mark | Guidance |
|---|---|---|
| At rest $\Rightarrow v = 0$ | M1 | $v=0$ seen or implied |
| $\Rightarrow \cos\theta = -\frac{3}{4}$ | A1 | Correct equation in $\theta$ |
| Acceleration is tangential | M1 | Correct direction for acceleration |
| Magnitude $\left|g\cos(\theta - 90)\right| = 6.48\ \text{m s}^{-2}$ or $\frac{\sqrt{7}}{4}g$ | A1 | Accept 6.48, 6.5 or exact in $g$ |
| At $\left(\cos^{-1}\left(-\frac{3}{4}\right) - 90 =\right) 48.6°$ to the downward vertical | A1 | Accept 0.848 (radians) |
**(5 marks) (14 marks total)**
---\begin{enumerate}
\item A small bead $B$ of mass $m$ is threaded on a circular hoop.
\end{enumerate}
The hoop has centre $O$ and radius $a$ and is fixed in a vertical plane.\\
The bead is projected with speed $\sqrt { \frac { 7 } { 2 } g a }$ from the lowest point of the hoop.\\
The hoop is modelled as being smooth.\\
When the angle between $O B$ and the downward vertical is $\theta$, the speed of $B$ is $v$.\\
(a) Show that $v ^ { 2 } = g a \left( \frac { 3 } { 2 } + 2 \cos \theta \right)$\\
(b) Find the size of $\theta$ at the instant when the contact force between $B$ and the hoop is first zero.\\
(c) Give a reason why your answer to part (b) is not likely to be the actual value of $\theta$.\\
(d) Find the magnitude and direction of the acceleration of $B$ at the instant when $B$ is first at instantaneous rest.
\hfill \mbox{\textit{Edexcel FM2 Q6 [14]}}