| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2008 |
| Session | January |
| Marks | 15 |
| Paper | Download PDF ↗ |
| Mark scheme | Download PDF ↗ |
| Topic | Circular Motion 2 |
| Type | Vertical circle: string becomes slack |
| Difficulty | Challenging +1.2 This is a standard M3 vertical circle problem with multiple parts requiring energy conservation, circular motion equations, and finding when string becomes slack. Part (a) is a routine 'show that' using energy conservation. Parts (b-c) apply standard T = mv²/r - mg cos θ and set T=0. Part (d) adds a peg complication requiring careful energy tracking but uses the same principles. More steps than average but all techniques are standard M3 material with no novel insight required. |
| Spec | 3.02h Motion under gravity: vector form6.02i Conservation of energy: mechanical energy principle6.05d Variable speed circles: energy methods |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| Energy equation with two terms on RHS: \(\frac{1}{2}mv^2 = \frac{1}{2}m\cdot\frac{5ga}{2} + mga\sin\theta\) | M1, A1 | |
| \(\Rightarrow v^2 = \frac{ga}{2}(5 + 4\sin\theta)\) | A1 cso | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(R(\ | \ \text{string})\ T - mg\sin\theta = \frac{mv^2}{a}\) (3 terms) | M1 A1 |
| \(\Rightarrow T = \frac{mg}{2}(5 + 6\sin\theta)\) o.e. | A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(T = 0 \Rightarrow \sin\theta = -\frac{5}{6}\) | M1, A1 | |
| Has a solution, so string slack when \(\alpha \approx 236(.4)°\) or \(4.13\) radians | A1 | (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| At top of small circle: \(\frac{1}{2}mv^2 = \frac{1}{2}m\cdot\frac{5ga}{2} - \frac{mga}{2}\) (M1 for energy equation with 3 terms) | M1 A1 | |
| \(\Rightarrow v^2 = \frac{3}{2}ga \approx 14.7a\) | A1 | |
| Resolving and using Force \(= \frac{mv^2}{r}\): \(T + mg = m\cdot\frac{\frac{3}{2}ga}{\frac{1}{2}a}\) (M1 needs three terms, but any \(v\)) | M1 A1 | |
| \(\Rightarrow T = 2mg\) | A1 | (6) |
## Question 6:
### Part (a):
| Answer/Working | Marks | Guidance |
|---|---|---|
| Energy equation with two terms on RHS: $\frac{1}{2}mv^2 = \frac{1}{2}m\cdot\frac{5ga}{2} + mga\sin\theta$ | M1, A1 | |
| $\Rightarrow v^2 = \frac{ga}{2}(5 + 4\sin\theta)$ | A1 cso | (3) |
### Part (b):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $R(\|\ \text{string})\ T - mg\sin\theta = \frac{mv^2}{a}$ (3 terms) | M1 A1 | |
| $\Rightarrow T = \frac{mg}{2}(5 + 6\sin\theta)$ o.e. | A1 | (3) |
### Part (c):
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T = 0 \Rightarrow \sin\theta = -\frac{5}{6}$ | M1, A1 | |
| Has a solution, so string slack when $\alpha \approx 236(.4)°$ or $4.13$ radians | A1 | (3) |
### Part (d):
| Answer/Working | Marks | Guidance |
|---|---|---|
| At top of small circle: $\frac{1}{2}mv^2 = \frac{1}{2}m\cdot\frac{5ga}{2} - \frac{mga}{2}$ (M1 for energy equation with 3 terms) | M1 A1 | |
| $\Rightarrow v^2 = \frac{3}{2}ga \approx 14.7a$ | A1 | |
| Resolving and using Force $= \frac{mv^2}{r}$: $T + mg = m\cdot\frac{\frac{3}{2}ga}{\frac{1}{2}a}$ (M1 needs three terms, but any $v$) | M1 A1 | |
| $\Rightarrow T = 2mg$ | A1 | (6) |
**Note:** Use of $v^2 = u^2 + 2gh$ is M0 in part (a).
**Total: 15 marks**6.
\begin{figure}[h]
\begin{center}
\captionsetup{labelformat=empty}
\caption{Figure 2}
\includegraphics[alt={},max width=\textwidth]{39c2d25a-a39b-4eb9-a17b-6e741ab5ae98-09_357_606_315_717}
\end{center}
\end{figure}
A particle $P$ of mass $m$ is attached to one end of a light inextensible string of length $a$. The other end of the string is attached to a fixed point $O$. At time $t = 0 , P$ is projected vertically downwards with speed $\sqrt { } \left( \frac { 5 } { 2 } g a \right)$ from a point $A$ which is at the same level as $O$ and a distance $a$ from $O$. When the string has turned through an angle $\theta$ and the string is still taut, the speed of $P$ is $v$ and the tension in the string is $T$, as shown in Figure 2.
\begin{enumerate}[label=(\alph*)]
\item Show that $v ^ { 2 } = \frac { g a } { 2 } ( 5 + 4 \sin \theta )$.
\item Find $T$ in terms of $m , g$ and $\theta$.
The string becomes slack when $\theta = \alpha$.
\item Find the value of $\alpha$.
The particle is projected again from $A$ with the same velocity as before. When $P$ is at the same level as $O$ for the first time after leaving $A$, the string meets a small smooth peg $B$ which has been fixed at a distance $\frac { 1 } { 2 } a$ from $O$. The particle now moves on an arc of a circle centre $B$. Given that the particle reaches the point $C$, which is $\frac { 1 } { 2 } a$ vertically above the point $B$, without the string going slack,
\item find the tension in the string when $P$ is at the point $C$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2008 Q6 [15]}}