| Exam Board | Edexcel |
|---|---|
| Module | FM2 (Further Mechanics 2) |
| Year | 2023 |
| Session | June |
| Marks | 9 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 1 |
| Type | Vertical circle – surface contact (sphere/track, leaving surface) |
| Difficulty | Standard +0.3 This is a standard vertical circular motion problem requiring energy conservation and Newton's second law in the radial direction. Part (a) is a guided 'show that' requiring two standard equations, and part (b) follows directly by setting R=0. While it's Further Maths content, it's a textbook application with no novel insight required, making it slightly easier than average overall. |
| Spec | 6.02i 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 | 3.1a - Dimensionally correct equation; all relevant terms; condone sign errors and sin/cos confusion |
| \(\frac{1}{2}Mv^2 = Mga(1-\cos\theta)\) | A1 | 1.1b - Correct unsimplified equation |
| Equation of motion (radial) | M1 | 3.1a - Dimensionally correct equation; all relevant terms; condone sign errors and sin/cos confusion |
| \(\frac{Mv^2}{a} = Mg\cos\theta - R\) | A1 | 1.1b - Correct unsimplified equation |
| Solve for \(R\) (complete method) | DM1 | 2.1 - If more than 2 equations, mark correct ones; if incorrect equation used then DM0 |
| \(R = Mg\cos\theta - \frac{Mv^2}{a} = Mg\cos\theta - 2Mg(1-\cos\theta) = 3Mg\cos\theta - 2Mg = Mg(3\cos\theta - 2)\) * | A1* | 2.2a - Obtain given answer from full and correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| \(R = 0 \Rightarrow \cos\theta = \frac{2}{3}\) | B1 | 1.1b - Seen or implied |
| \(v^2 = 2ga(1-\cos\theta)\) | M1 | 3.1a |
| \(v^2 = \frac{2}{3}ga\), \(v = \sqrt{\frac{2ga}{3}}\) | A1 | 1.1b |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int \frac{1}{2}y^2 dx = \frac{1}{2}\int(9-x^2)^2 dx = \frac{1}{2}\int(81-18x^2+x^4)dx\) | M1 | Use of \(\int \frac{1}{2}y^2 dx\) or \(\int xy\, dy\). Ignore any limits |
| \(= \frac{1}{2}\left[81x - 6x^3 + \frac{1}{5}x^5\right]_0^3\) | DM1 | Integrate and use correct limits (0 and 3 for \(x\), 0 and 9 for \(y\)). Powers increasing by 1 |
| \(= \frac{1}{2}\left(3\times81 - 6\times27 + \frac{243}{5}\right) = \frac{324}{5}\) | A1 | Correct unsimplified expression |
| \(\Rightarrow \text{distance} = \frac{64.8}{18} = 3.6\) * | A1* | Obtain given answer from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\int xy\,dy = \int y\sqrt{9-y}\,dy\) | M1 | 2.1 |
| \(= \left[-\frac{2}{3}y(9-y)^{\frac{3}{2}} - \frac{4}{15}(9-y)^{\frac{5}{2}}\right]_0^9\) | DM1 | 1.1b |
| \(= \frac{4}{15}\times 9^{\frac{5}{2}} = \frac{324}{5}\) | A1 | 1.1b |
| \(\Rightarrow \text{distance} = \frac{64.8}{18} = 3.6\text{(m)}\) * | A1* | 2.2a |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Moments about \(O\) | M1 | Complete method to obtain \(\lambda\), e.g. by taking moments about \(O\) or by resolving and taking moments about a different point. Allow for an equation in \(T_A\) or \(\lambda\) |
| \(3.6W = 9\lambda W\) | A1 | Correct unsimplified equation in \(\lambda\) |
| \(\lambda = 0.4\) | A1 | Correct only |
# Question 4(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Conservation of energy | M1 | 3.1a - Dimensionally correct equation; all relevant terms; condone sign errors and sin/cos confusion |
| $\frac{1}{2}Mv^2 = Mga(1-\cos\theta)$ | A1 | 1.1b - Correct unsimplified equation |
| Equation of motion (radial) | M1 | 3.1a - Dimensionally correct equation; all relevant terms; condone sign errors and sin/cos confusion |
| $\frac{Mv^2}{a} = Mg\cos\theta - R$ | A1 | 1.1b - Correct unsimplified equation |
| Solve for $R$ (complete method) | DM1 | 2.1 - If more than 2 equations, mark correct ones; if incorrect equation used then DM0 |
| $R = Mg\cos\theta - \frac{Mv^2}{a} = Mg\cos\theta - 2Mg(1-\cos\theta) = 3Mg\cos\theta - 2Mg = Mg(3\cos\theta - 2)$ * | A1* | 2.2a - Obtain given answer from full and correct working |
---
# Question 4(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| $R = 0 \Rightarrow \cos\theta = \frac{2}{3}$ | B1 | 1.1b - Seen or implied |
| $v^2 = 2ga(1-\cos\theta)$ | M1 | 3.1a |
| $v^2 = \frac{2}{3}ga$, $v = \sqrt{\frac{2ga}{3}}$ | A1 | 1.1b |
## Question 5a:
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int \frac{1}{2}y^2 dx = \frac{1}{2}\int(9-x^2)^2 dx = \frac{1}{2}\int(81-18x^2+x^4)dx$ | M1 | Use of $\int \frac{1}{2}y^2 dx$ or $\int xy\, dy$. Ignore any limits |
| $= \frac{1}{2}\left[81x - 6x^3 + \frac{1}{5}x^5\right]_0^3$ | DM1 | Integrate and use correct limits (0 and 3 for $x$, 0 and 9 for $y$). Powers increasing by 1 |
| $= \frac{1}{2}\left(3\times81 - 6\times27 + \frac{243}{5}\right) = \frac{324}{5}$ | A1 | Correct unsimplified expression |
| $\Rightarrow \text{distance} = \frac{64.8}{18} = 3.6$ * | A1* | Obtain given answer from correct working |
**Alternative (5a alt):**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\int xy\,dy = \int y\sqrt{9-y}\,dy$ | M1 | 2.1 |
| $= \left[-\frac{2}{3}y(9-y)^{\frac{3}{2}} - \frac{4}{15}(9-y)^{\frac{5}{2}}\right]_0^9$ | DM1 | 1.1b |
| $= \frac{4}{15}\times 9^{\frac{5}{2}} = \frac{324}{5}$ | A1 | 1.1b |
| $\Rightarrow \text{distance} = \frac{64.8}{18} = 3.6\text{(m)}$ * | A1* | 2.2a |
---
## Question 5b:
| Answer/Working | Mark | Guidance |
|---|---|---|
| Moments about $O$ | M1 | Complete method to obtain $\lambda$, e.g. by taking moments about $O$ or by resolving and taking moments about a different point. Allow for an equation in $T_A$ or $\lambda$ |
| $3.6W = 9\lambda W$ | A1 | Correct unsimplified equation in $\lambda$ |
| $\lambda = 0.4$ | A1 | Correct only |
---4.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{3b070338-1de4-4c33-be29-d37ac06c9fed-12_490_1177_219_507}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\end{center}
\end{figure}
A smooth hemisphere of radius $a$ is fixed on a horizontal surface with its plane face in contact with the surface. The centre of the plane face of the hemisphere is $O$.
A particle $P$ of mass $M$ is disturbed from rest at the highest point of the hemisphere.\\
When $P$ is still on the surface of the hemisphere and the radius from $O$ to $P$ is at an angle $\theta$ to the vertical,
\begin{itemize}
\item the speed of $P$ is $v$
\item the normal reaction between the hemisphere and the particle is $R$, as shown in Figure 2.
\begin{enumerate}[label=(\alph*)]
\item Show that $\mathrm { R } = \mathrm { Mg } ( 3 \cos \theta - 2 )$
\item Find, in terms of $a$ and $g$, the speed of the particle at the instant when the particle leaves the surface of the hemisphere.
\end{itemize}
\end{enumerate}
\hfill \mbox{\textit{Edexcel FM2 2023 Q4 [9]}}