| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2018 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: string becomes slack |
| Difficulty | Standard +0.8 This is a multi-part vertical circle problem requiring energy conservation, circular motion dynamics (centripetal force), and projectile motion after the string slackens. Part (a) is a standard 'show that' requiring energy and Newton's second law in circular motion. Parts (b) and (c) require recognizing that the string becomes slack when T=0, then applying projectile motion with careful geometry. The problem demands multiple techniques and careful reasoning about the transition between constrained and free motion, placing it moderately above average difficulty. |
| Spec | 3.02h Motion under gravity: vector form6.02e Calculate KE and PE: using formulae6.02i Conservation of energy: mechanical energy principle6.05d Variable speed circles: energy methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\frac{1}{2}mv^2 - \frac{1}{2}m\times 2lg = mgl\cos\theta\) | M1, A1 | Forming an energy equation from start to general position. 2 KE terms needed and a change in PE (with 1 or 2 terms). Correct equation |
| \(T - mg\cos\theta = m\frac{v^2}{l}\) | M1, A1, A1 | Attempting equation of motion along the radius. Weight must be resolved. Correct difference of forces. Acceleration as shown |
| \(T = mg\cos\theta + \frac{1}{l}(2mgl\cos\theta + 2mgl)\) | DM1 | Eliminate \(v^2\) between the 2 equations. Depends on both previous M marks |
| \(T = mg(3\cos\theta + 2)\) | A1cso (7) | Obtain the given expression with no errors |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(T = 0 \Rightarrow \cos\theta = -\frac{2}{3}\) | B1 | \(\cos\theta = -\frac{2}{3}\) at point where string becomes slack |
| \(v^2 = -gl\cos\theta\) or \(v^2 = 2gl\cos\theta + 2gl\) | ||
| \(v^2 = \frac{2gl}{3},\quad v = \sqrt{\frac{2gl}{3}}\) | M1, A1 (3) | Obtaining \(v^2\) or \(v\) in terms of \(g\) and \(l\). Correct expression for \(v\), any equivalent form |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Horiz speed at \(B = v | \cos\theta | = \sqrt{\frac{2gl}{3}}\times\frac{2}{3}\) |
| Energy: \(\frac{1}{2}m\left(\frac{2gl}{3}\right) - \frac{1}{2}m\times\frac{4}{9}\left(\frac{2gl}{3}\right) = mgh\) | M1 | Attempting energy equation from \(B\) to highest point, 2 non-zero KE terms needed |
| \(h = \frac{5l}{27}\) | A1 | Obtain correct height above \(B\) |
| Height above \(O = \frac{5l}{27} + l\cos(180-\theta) = \frac{5l}{27}+\frac{2l}{3} = \frac{23l}{27}\) (0.85\(l\) or 0.852\(l\)) | A1 cao (5) | Complete to correct height above \(O\) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Vert speed at \(B = v\cos(\theta-90) = v\sin\theta = \sqrt{\frac{2gl}{3}}\times\frac{\sqrt{5}}{3}\) | M1, A1ft on \(v\) | Resolve speed at \(B\) to obtain vertical component. \(\frac{\sqrt{5}}{3}\times\) their speed |
| \(0 = \frac{2gl}{3}\times\frac{5}{9} - 2gs\) | M1 | Use \(v^2 = u^2 + 2as\) with their (non-zero) \(u\) and \(v=0\) to obtain height above \(B\) |
| \(s = \frac{5l}{27}\) | A1 | Correct height above \(B\) |
| Height above \(O = \frac{5l}{27}+l\cos(180-\theta) = \frac{5l}{27}+\frac{2l}{3} = \frac{23l}{27}\) | A1 (5) | Complete to correct height above \(O\) |
## Question 6(a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\frac{1}{2}mv^2 - \frac{1}{2}m\times 2lg = mgl\cos\theta$ | M1, A1 | Forming an energy equation from start to general position. 2 KE terms needed and a change in PE (with 1 or 2 terms). Correct equation |
| $T - mg\cos\theta = m\frac{v^2}{l}$ | M1, A1, A1 | Attempting equation of motion along the radius. Weight must be resolved. Correct difference of forces. Acceleration as shown |
| $T = mg\cos\theta + \frac{1}{l}(2mgl\cos\theta + 2mgl)$ | DM1 | Eliminate $v^2$ between the 2 equations. Depends on both previous M marks |
| $T = mg(3\cos\theta + 2)$ | A1cso (7) | Obtain the given expression with no errors |
---
## Question 6(b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T = 0 \Rightarrow \cos\theta = -\frac{2}{3}$ | B1 | $\cos\theta = -\frac{2}{3}$ at point where string becomes slack |
| $v^2 = -gl\cos\theta$ or $v^2 = 2gl\cos\theta + 2gl$ | | |
| $v^2 = \frac{2gl}{3},\quad v = \sqrt{\frac{2gl}{3}}$ | M1, A1 (3) | Obtaining $v^2$ or $v$ in terms of $g$ and $l$. Correct expression for $v$, any equivalent form |
---
## Question 6(c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Horiz speed at $B = v|\cos\theta| = \sqrt{\frac{2gl}{3}}\times\frac{2}{3}$ | M1, A1ft on $v$ | Resolve speed at $B$ to obtain horizontal component. $\frac{2}{3}\times$ their speed |
| Energy: $\frac{1}{2}m\left(\frac{2gl}{3}\right) - \frac{1}{2}m\times\frac{4}{9}\left(\frac{2gl}{3}\right) = mgh$ | M1 | Attempting energy equation from $B$ to highest point, 2 non-zero KE terms needed |
| $h = \frac{5l}{27}$ | A1 | Obtain correct height above $B$ |
| Height above $O = \frac{5l}{27} + l\cos(180-\theta) = \frac{5l}{27}+\frac{2l}{3} = \frac{23l}{27}$ (0.85$l$ or 0.852$l$) | A1 cao (5) | Complete to correct height above $O$ |
**ALT(c):**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Vert speed at $B = v\cos(\theta-90) = v\sin\theta = \sqrt{\frac{2gl}{3}}\times\frac{\sqrt{5}}{3}$ | M1, A1ft on $v$ | Resolve speed at $B$ to obtain vertical component. $\frac{\sqrt{5}}{3}\times$ their speed |
| $0 = \frac{2gl}{3}\times\frac{5}{9} - 2gs$ | M1 | Use $v^2 = u^2 + 2as$ with their (non-zero) $u$ and $v=0$ to obtain height above $B$ |
| $s = \frac{5l}{27}$ | A1 | Correct height above $B$ |
| Height above $O = \frac{5l}{27}+l\cos(180-\theta) = \frac{5l}{27}+\frac{2l}{3} = \frac{23l}{27}$ | A1 (5) | Complete to correct height above $O$ |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{d93ae982-9395-4311-9972-be727b3ce954-18_483_730_242_609}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}
A particle $P$ of mass $m$ is attached to one end of a light inextensible string of length $l$. The other end of the string is attached to a fixed point $O$. The particle is held at the point $A$, where $O A = l$ and $O A$ is horizontal. The particle is then projected vertically downwards from $A$ with speed $\sqrt { 2 g l }$, as shown in Figure 4 . When the string makes an angle $\theta$ with the downward vertical through $O$ and the string is still taut, the tension in the string is $T$.
\begin{enumerate}[label=(\alph*)]
\item Show that $T = m g ( 3 \cos \theta + 2 )$
At the instant when the particle reaches the point $B$, the string becomes slack.
\item Find the speed of $P$ at $B$.
\item Find the greatest height above $O$ reached by $P$ in the subsequent motion.
\begin{center}
\end{center}
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2018 Q6 [15]}}