| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2024 |
| Session | June |
| Marks | 13 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Particle on outer surface of sphere |
| Difficulty | Standard +0.8 This is a challenging M3 circular motion problem requiring multiple sophisticated steps: applying conservation of energy, using the condition for loss of contact (N=0 implies centripetal force equals weight component), then projectile motion with careful geometry. The algebra is non-trivial and requires connecting three distinct phases of motion. Significantly harder than standard circular motion questions but within reach of strong M3 students. |
| Spec | 3.02h Motion under gravity: vector form6.02i Conservation of energy: mechanical energy principle6.05e Radial/tangential acceleration6.05f Vertical circle: motion including free fall |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\frac{1}{2}mv^2 - \frac{1}{2}mu^2 = mgr(1-\cos\theta)\) | M1A2 | Use conservation of energy. Dimensionally correct, all terms present, no extras. Condone sign errors and sin/cos confusion. |
| \(mg\cos\theta = \dfrac{mv^2}{r}\) | M1A1 | Use N2L to form equation of motion towards \(O\). \(R=0\) must be used at some point. Allow with or without \(R\). |
| Eliminate \(v^2\) and solve for \(\cos\theta\) | M1 | |
| \(\cos\theta = \dfrac{2gr+u^2}{3gr}\) * | A1* | Given answer from complete and correct working. Must include a line with \(v^2\) eliminated before reaching given answer. |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(\cos\theta = \dfrac{4}{5}\) | B1 | Seen or implied |
| \(v^2 = rg\cos\theta \quad \left(v = \sqrt{\dfrac{4rg}{5}}\right)\) | M1 | Solve to find \(v\) in terms of \(g\), \(r\) and \(\theta\) |
| Horiz component at \(C\): \(H = v\cos\theta\), \(\left(H = \dfrac{4}{5}\sqrt{\dfrac{4rg}{5}} = \sqrt{\dfrac{64rg}{125}}\right)\) | M1 | Correct method to find at least one of \(H\), \(V\) or \(W\). Condone finding \(H^2\), \(V^2\) or \(W^2\). Dimensionally correct, all terms present. M0 if speed from (a) used instead of \(v\). |
| Vert component at \(C\): \(V = \sqrt{(v\sin\theta)^2 + 2gr\cos\theta}\), \(\left(V = \sqrt{\dfrac{236rg}{125}}\right)\) | M1 | Correct method to find any two of \(H\), \(V\) or \(W\). Same conditions as above. |
| Speed at \(C\): \(W\) where \(\frac{1}{2}m\left(W^2 - \frac{2gr}{5}\right) = mgr\) OR \(\frac{1}{2}m(W^2-v^2)=mgr\cos\theta\), \(\left(W=\sqrt{\dfrac{12rg}{5}}\right)\) | ||
| \(\tan\alpha = \dfrac{V}{H} = \dfrac{\sqrt{W^2-H^2}}{H} = \dfrac{V}{\sqrt{W^2-V^2}}\) | DM1 | Dependent on previous two M marks. Complete method to find \(\tan\alpha\). Condone if they go straight to \(\alpha = \tan^{-1}(\ldots)\) and never state \(\tan\alpha =\) |
| \(\tan\alpha = \dfrac{\sqrt{59}}{4}\) | A1 | Correct value. Accept equivalent surds e.g. \(\sqrt{\dfrac{59}{16}}\). Must be exact. A0 if they go straight to \(\alpha\) and never find \(\tan\alpha\). |
## Question 6:
### Part 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}mv^2 - \frac{1}{2}mu^2 = mgr(1-\cos\theta)$ | M1A2 | Use conservation of energy. Dimensionally correct, all terms present, no extras. Condone sign errors and sin/cos confusion. |
| $mg\cos\theta = \dfrac{mv^2}{r}$ | M1A1 | Use N2L to form equation of motion towards $O$. $R=0$ must be used at some point. Allow with or without $R$. |
| Eliminate $v^2$ and solve for $\cos\theta$ | M1 | |
| $\cos\theta = \dfrac{2gr+u^2}{3gr}$ * | A1* | Given answer from complete and correct working. Must include a line with $v^2$ eliminated before reaching given answer. |
### Part 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\cos\theta = \dfrac{4}{5}$ | B1 | Seen or implied |
| $v^2 = rg\cos\theta \quad \left(v = \sqrt{\dfrac{4rg}{5}}\right)$ | M1 | Solve to find $v$ in terms of $g$, $r$ and $\theta$ |
| Horiz component at $C$: $H = v\cos\theta$, $\left(H = \dfrac{4}{5}\sqrt{\dfrac{4rg}{5}} = \sqrt{\dfrac{64rg}{125}}\right)$ | M1 | Correct method to find at least one of $H$, $V$ or $W$. Condone finding $H^2$, $V^2$ or $W^2$. Dimensionally correct, all terms present. M0 if speed from (a) used instead of $v$. |
| Vert component at $C$: $V = \sqrt{(v\sin\theta)^2 + 2gr\cos\theta}$, $\left(V = \sqrt{\dfrac{236rg}{125}}\right)$ | M1 | Correct method to find any two of $H$, $V$ or $W$. Same conditions as above. |
| Speed at $C$: $W$ where $\frac{1}{2}m\left(W^2 - \frac{2gr}{5}\right) = mgr$ **OR** $\frac{1}{2}m(W^2-v^2)=mgr\cos\theta$, $\left(W=\sqrt{\dfrac{12rg}{5}}\right)$ | | |
| $\tan\alpha = \dfrac{V}{H} = \dfrac{\sqrt{W^2-H^2}}{H} = \dfrac{V}{\sqrt{W^2-V^2}}$ | DM1 | Dependent on previous two M marks. Complete method to find $\tan\alpha$. Condone if they go straight to $\alpha = \tan^{-1}(\ldots)$ and never state $\tan\alpha =$ |
| $\tan\alpha = \dfrac{\sqrt{59}}{4}$ | A1 | Correct value. Accept equivalent surds e.g. $\sqrt{\dfrac{59}{16}}$. Must be exact. A0 if they go straight to $\alpha$ and never find $\tan\alpha$. |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{176ae8f8-7de9-4993-825a-6067614526ae-16_739_921_299_699}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
A fixed solid sphere has centre $O$ and radius $r$.\\
A particle $P$ of mass $m$ is held at rest on the smooth surface of the sphere at $A$, the highest point of the sphere.\\
The particle $P$ is then projected horizontally from $A$ with speed $u$ and moves on the surface of the sphere.\\
At the instant when $P$ reaches the point $B$ on the sphere, where angle $A O B = \theta , P$ is moving with speed $v$, as shown in Figure 4.
At this instant, $P$ loses contact with the surface of the sphere.
\begin{enumerate}[label=(\alph*)]
\item Show that
$$\cos \theta = \frac { 2 g r + u ^ { 2 } } { 3 g r }$$
In the subsequent motion, the particle $P$ crosses the horizontal through $O$ at the point $C$, also shown in Figure 4.
At the instant $P$ passes through $C , P$ is moving at an angle $\alpha$ to the horizontal.\\
Given that $u ^ { 2 } = \frac { 2 g r } { 5 }$
\item find the exact value of $\tan \alpha$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2024 Q6 [13]}}