| Exam Board | Edexcel |
|---|---|
| Module | FM2 (Further Mechanics 2) |
| Year | 2024 |
| Session | June |
| Marks | 14 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Vertical circle – surface contact (sphere/track, leaving surface) |
| Difficulty | Challenging +1.2 This is a standard Further Maths vertical circular motion problem requiring energy conservation and Newton's second law in the radial direction. Part (a) is routine energy conservation, (b) requires finding the leaving condition (N=0), (c) uses energy again, and (d) needs velocity components. While it's multi-part and requires careful algebra, all techniques are standard FM2 material with no novel insights required. Slightly above average difficulty due to the algebraic manipulation and multi-step nature, but well within expected FM2 scope. |
| 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 |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Energy equation | M1 | Conservation of mechanical energy. All terms required, no extras; dimensionally correct. Condone sine/cosine confusion and sign errors. |
| \(\frac{1}{2}mU^2 + mgr(1-\cos\theta) = \frac{1}{2}mv^2\) | A1 | Correct unsimplified equation |
| \(v^2 = U^2 + 2gr(1-\cos\theta)\) | A1 | Or equivalent with \(v^2\) as subject |
| (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Equation for circular motion at \(B\) | M1 | Dimensionally correct. Condone sine/cosine confusion and sign errors. Condone if \(R=0\) seen or implied at this stage. |
| \(mg\cos\theta - R = \frac{mW^2}{r}\) | A1 | Correct unsimplified equation |
| Use \(R=0\): \(mg\cos\theta = \frac{m(U^2 + 2gr(1-\cos\theta))}{r}\) | M1 | Use \(R=0\) in a relevant equation to obtain equation in \(r\), \(g\) and \(\cos\theta\) or \(W\) |
| \(rmg\cos\theta = m\left(\frac{2rg}{3} + 2gr(1-\cos\theta)\right) \Rightarrow \cos\theta = \frac{8}{9}\) | A1 | Any correct equation in \(r\), \(g\) and \(\cos\theta\) or \(r\), \(g\) and \(W\). E.g. \(W^2 = \frac{2gr}{3} + 2gr - 2W^2\) |
| \(\Rightarrow W^2 = rg\cos\theta = \frac{8}{9}rg\) | A1* | Obtain given result from correct exact working |
| (5 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Energy equation | M1 | Complete method to find the speed, e.g. by using conservation of energy or projectile motion. All terms required, no extras; dimensionally correct. |
| From \(B\): \(\frac{1}{2}mV^2 = \frac{1}{2}mW^2 + mgr\cos\theta\) or from \(A\): \(\frac{1}{2}mV^2 = \frac{1}{2}mU^2 + mgr\) | A1 | Correct unsimplified equation(s) |
| \(V^2 = v^2 + 2gr\cos\theta = \frac{8rg}{9} + \frac{16rg}{9} = \frac{8rg}{3}\), \(V = \sqrt{\frac{8rg}{3}}\) | A1 | Any equivalent form |
| (3 marks) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Form an equation in \(\alpha\) | M1 | Complete method to form a trig ratio for \(\alpha\) |
| \(\cos\alpha° = \frac{W\cos\theta}{V} = \frac{\frac{8}{9}\sqrt{\frac{8rg}{9}}}{\sqrt{\frac{8rg}{3}}} = \frac{8\sqrt{3}}{27}\) | A1ft | Correct use of their values to obtain a ratio for \(\alpha\). Ft their \(V\). \(\left(\sin\alpha = \frac{\sqrt{537}}{27} = 0.858,\ \tan\alpha = \frac{\sqrt{179}}{8} = 1.67\right)\) |
| \(\alpha = 59(.12276...)\) | A1 | 59 or better |
| (3 marks) | ||
| Total: 14 marks |
## Question 7:
### Part 7a:
| Answer/Working | Marks | Guidance |
|---|---|---|
| Energy equation | M1 | Conservation of mechanical energy. All terms required, no extras; dimensionally correct. Condone sine/cosine confusion and sign errors. |
| $\frac{1}{2}mU^2 + mgr(1-\cos\theta) = \frac{1}{2}mv^2$ | A1 | Correct unsimplified equation |
| $v^2 = U^2 + 2gr(1-\cos\theta)$ | A1 | Or equivalent with $v^2$ as subject |
| **(3 marks)** | | |
### Part 7b:
| Answer/Working | Marks | Guidance |
|---|---|---|
| Equation for circular motion at $B$ | M1 | Dimensionally correct. Condone sine/cosine confusion and sign errors. Condone if $R=0$ seen or implied at this stage. |
| $mg\cos\theta - R = \frac{mW^2}{r}$ | A1 | Correct unsimplified equation |
| Use $R=0$: $mg\cos\theta = \frac{m(U^2 + 2gr(1-\cos\theta))}{r}$ | M1 | Use $R=0$ in a relevant equation to obtain equation in $r$, $g$ and $\cos\theta$ or $W$ |
| $rmg\cos\theta = m\left(\frac{2rg}{3} + 2gr(1-\cos\theta)\right) \Rightarrow \cos\theta = \frac{8}{9}$ | A1 | Any correct equation in $r$, $g$ and $\cos\theta$ or $r$, $g$ and $W$. E.g. $W^2 = \frac{2gr}{3} + 2gr - 2W^2$ |
| $\Rightarrow W^2 = rg\cos\theta = \frac{8}{9}rg$ | A1* | Obtain given result from correct exact working |
| **(5 marks)** | | |
### Part 7c:
| Answer/Working | Marks | Guidance |
|---|---|---|
| Energy equation | M1 | **Complete** method to find the speed, e.g. by using conservation of energy or projectile motion. All terms required, no extras; dimensionally correct. |
| From $B$: $\frac{1}{2}mV^2 = \frac{1}{2}mW^2 + mgr\cos\theta$ or from $A$: $\frac{1}{2}mV^2 = \frac{1}{2}mU^2 + mgr$ | A1 | Correct unsimplified equation(s) |
| $V^2 = v^2 + 2gr\cos\theta = \frac{8rg}{9} + \frac{16rg}{9} = \frac{8rg}{3}$, $V = \sqrt{\frac{8rg}{3}}$ | A1 | Any equivalent form |
| **(3 marks)** | | |
### Part 7d:
| Answer/Working | Marks | Guidance |
|---|---|---|
| Form an equation in $\alpha$ | M1 | Complete method to form a trig ratio for $\alpha$ |
| $\cos\alpha° = \frac{W\cos\theta}{V} = \frac{\frac{8}{9}\sqrt{\frac{8rg}{9}}}{\sqrt{\frac{8rg}{3}}} = \frac{8\sqrt{3}}{27}$ | A1ft | Correct use of their values to obtain a ratio for $\alpha$. Ft their $V$. $\left(\sin\alpha = \frac{\sqrt{537}}{27} = 0.858,\ \tan\alpha = \frac{\sqrt{179}}{8} = 1.67\right)$ |
| $\alpha = 59(.12276...)$ | A1 | 59 or better |
| **(3 marks)** | | |
| **Total: 14 marks** | | |7.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{c14975b7-6afa-44ce-beab-1cba2e82b249-24_419_935_251_566}
\captionsetup{labelformat=empty}
\caption{Figure 6}
\end{center}
\end{figure}
A smooth solid hemisphere has radius $r$ and the centre of its plane face is $O$.\\
The hemisphere is fixed with its plane face in contact with horizontal ground, as shown in Figure 6.\\
A small stone is at the point $A$, the highest point on the surface of the hemisphere. The stone is projected horizontally from $A$ with speed $U$.\\
The stone is still in contact with the hemisphere at the point $B$, where $O B$ makes an angle $\theta$ with the upward vertical.\\
The speed of the stone at the instant it reaches $B$ is $v$.\\
The stone is modelled as a particle $P$ and air resistance is modelled as being negligible.
\begin{enumerate}[label=(\alph*)]
\item Use the model to find $v ^ { 2 }$ in terms of $U , r , g$ and $\theta$
When $P$ leaves the surface of the hemisphere, the speed of $P$ is $W$.\\
Given that $U = \sqrt { \frac { 2 r g } { 3 } }$
\item show that $W ^ { 2 } = \frac { 8 } { 9 } r g$
After leaving the surface of the hemisphere, $P$ moves freely under gravity until it hits the ground.
\item Find the speed of $P$ as it hits the ground, giving your answer in terms of $r$ and $g$.
At the instant when $P$ hits the ground it is travelling at $\alpha ^ { \circ }$ to the horizontal.
\item Find the value of $\alpha$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FM2 2024 Q7 [14]}}