| Exam Board | Edexcel |
|---|---|
| Module | FM2 (Further Mechanics 2) |
| Year | 2019 |
| Session | June |
| Marks | 12 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Energy considerations in circular motion |
| Difficulty | Standard +0.8 This is a multi-part Further Mechanics question requiring energy conservation, understanding of circular motion dynamics (centripetal vs tangential acceleration), and force analysis. Part (a) tests conceptual understanding of instantaneous rest, parts (b) and (c) require applying energy methods and resolving forces in circular motion. While systematic, it demands careful coordination of multiple FM2 concepts and is more challenging than standard A-level mechanics. |
| Spec | 3.03d Newton's second law: 2D vectors6.02i Conservation of energy: mechanical energy principle6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(v = 0 \Rightarrow \frac{v^2}{L} = 0 \Rightarrow\) no acceleration towards \(O\) | B1 | Clear explanation using \(v = 0\) |
| \(\Rightarrow\) acceleration is perpendicular to \(OA\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Conservation of energy applied | M1 | All terms required. Must be dimensionally correct. Condone sign errors. \(v = 0\) seen or implied |
| \(0 = \frac{1}{2}mu^2 - mgL\cos\theta \quad \left(0 = \frac{2gL}{3} - 2gL\cos\theta\right)\) | A1 | Correct unsimplified equation for their \(\theta\) |
| \(\Rightarrow \cos\theta = \frac{1}{3}\) | A1 | Or equivalent to give trig ratio for relevant angle (taking account of their \(\theta\)) |
| Complete strategy to find the angle and \ | acceleration\ | |
| Magnitude: \(g\sin\theta = \frac{2\sqrt{2}}{3}g\) | A1 | Correct magnitude from correct work only. Accept 9.2, 9.24 |
| \(\theta = 71°\) or better | A1 | Correct value of \(\theta\) (1.2 radians or better) from correct work only |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Circular motion equation applied | M1 | Equation for circular motion. Need all terms and dimensionally correct. Condone sign errors. |
| \(T - mg = \frac{mv^2}{L}\) | A1 | Correct unsimplified equation |
| Energy equation applied | M1 | Use of conservation of energy. Require all 3 terms and dimensionally correct. |
| \(v^2 = \frac{2gL}{3} + 2gL \left(= \frac{8gL}{3}\right)\) | A1 | Correct unsimplified equation |
| \(T = mg + \frac{8mg}{3} = \frac{11mg}{3}\) | A1 | Correct only |
## Question 7:
### Part 7(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $v = 0 \Rightarrow \frac{v^2}{L} = 0 \Rightarrow$ no acceleration towards $O$ | B1 | Clear explanation using $v = 0$ |
| $\Rightarrow$ acceleration is perpendicular to $OA$ | | |
**Total: (1)**
---
### Part 7(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Conservation of energy applied | M1 | All terms required. Must be dimensionally correct. Condone sign errors. $v = 0$ seen or implied |
| $0 = \frac{1}{2}mu^2 - mgL\cos\theta \quad \left(0 = \frac{2gL}{3} - 2gL\cos\theta\right)$ | A1 | Correct unsimplified equation for their $\theta$ |
| $\Rightarrow \cos\theta = \frac{1}{3}$ | A1 | Or equivalent to give trig ratio for relevant angle (taking account of their $\theta$) |
| Complete strategy to find the angle and \|acceleration\| | M1 | Complete strategy to find our $\theta$ or magnitude of acceleration |
| Magnitude: $g\sin\theta = \frac{2\sqrt{2}}{3}g$ | A1 | Correct magnitude from correct work only. Accept 9.2, 9.24 |
| $\theta = 71°$ or better | A1 | Correct value of $\theta$ (1.2 radians or better) from correct work only |
**Total: (6)**
---
### Part 7(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Circular motion equation applied | M1 | Equation for circular motion. Need all terms and dimensionally correct. Condone sign errors. |
| $T - mg = \frac{mv^2}{L}$ | A1 | Correct unsimplified equation |
| Energy equation applied | M1 | Use of conservation of energy. Require all 3 terms and dimensionally correct. |
| $v^2 = \frac{2gL}{3} + 2gL \left(= \frac{8gL}{3}\right)$ | A1 | Correct unsimplified equation |
| $T = mg + \frac{8mg}{3} = \frac{11mg}{3}$ | A1 | Correct only |
**Total: (5)**
**Overall Total: (12 marks)**\begin{enumerate}
\item A particle, $P$, of mass $m$ is attached to one end of a light rod of length $L$. The other end of the rod is attached to a fixed point $O$ so that the rod is free to rotate in a vertical plane about $O$. The particle is held with the rod horizontal and is then projected vertically downwards with speed $u$. The particle first comes to instantaneous rest at the point $A$.\\
(a) Explain why the acceleration of $P$ at $A$ is perpendicular to $O A$.
\end{enumerate}
At the instant when $P$ is at the point $A$ the acceleration of $P$ is in a direction making an angle $\theta$ with the horizontal. Given that $u ^ { 2 } = \frac { 2 g L } { 3 }$,\\
(b) find\\
(i) the magnitude of the acceleration of $P$ at the point $A$,\\
(ii) the size of $\theta$.\\
(c) Find, in terms of $m$ and $g$, the magnitude of the tension in the rod at the instant when $P$ is at its lowest point.
\hfill \mbox{\textit{Edexcel FM2 2019 Q7 [12]}}