| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2006 |
| Session | January |
| Marks | 14 |
| 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, and projectile motion after the string slackens. Part (a) is a standard 'show that' using centripetal force, but parts (b) and (c) require finding when tension becomes zero, then analyzing subsequent projectile motion to find maximum height—this demands careful coordination of multiple mechanics concepts beyond routine exercises. |
| 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: \(\dfrac{1}{2}m(u^2 - v^2) = mgl(1 - \cos\theta)\) | M1 A1 | |
| \(\left[v^2 = gl + 2gl\cos\theta\right]\) | ||
| N2L: \(T - mg\cos\theta = \dfrac{mv^2}{l}\) | M1 A1 | |
| \(= \dfrac{mg(1 + 2\cos\theta)}{1}\) | M1 | |
| \(T = mg(1 + 3\cos\theta)\) ✱ | A1 | cso |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(T = 0 \Rightarrow \cos\theta = -\dfrac{1}{3}\) | B1 | |
| \(v^2 = gl - \dfrac{2}{3}gl \Rightarrow v = \left(\dfrac{gl}{3}\right)^{1/2}\) | M1 A1 |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Marks | Guidance |
| \(\uparrow\; v_y = \left(\dfrac{gl}{3}\right)^{1/2}\sin\theta \left[= \left(\dfrac{gl}{3}\right)^{1/2} \cdot \dfrac{2\sqrt{2}}{3}\right]\) | M1 | |
| \(v^2 = u^2 - 2gh \Rightarrow 2gh = \dfrac{gl}{3}\cdot\dfrac{8}{9} \Rightarrow h = \dfrac{4l}{27}\) | M1 A1 | |
| \(H = l(1 - \cos\theta) + \dfrac{4l}{27} = \dfrac{40l}{27}\) | M1 A1 |
## Question 6:
**Part (a)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| Energy: $\dfrac{1}{2}m(u^2 - v^2) = mgl(1 - \cos\theta)$ | M1 A1 | |
| $\left[v^2 = gl + 2gl\cos\theta\right]$ | | |
| N2L: $T - mg\cos\theta = \dfrac{mv^2}{l}$ | M1 A1 | |
| $= \dfrac{mg(1 + 2\cos\theta)}{1}$ | M1 | |
| $T = mg(1 + 3\cos\theta)$ ✱ | A1 | cso |
**(6 marks)**
**Part (b)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $T = 0 \Rightarrow \cos\theta = -\dfrac{1}{3}$ | B1 | |
| $v^2 = gl - \dfrac{2}{3}gl \Rightarrow v = \left(\dfrac{gl}{3}\right)^{1/2}$ | M1 A1 | |
**(3 marks)**
**Part (c)**
| Answer/Working | Marks | Guidance |
|---|---|---|
| $\uparrow\; v_y = \left(\dfrac{gl}{3}\right)^{1/2}\sin\theta \left[= \left(\dfrac{gl}{3}\right)^{1/2} \cdot \dfrac{2\sqrt{2}}{3}\right]$ | M1 | |
| $v^2 = u^2 - 2gh \Rightarrow 2gh = \dfrac{gl}{3}\cdot\dfrac{8}{9} \Rightarrow h = \dfrac{4l}{27}$ | M1 A1 | |
| $H = l(1 - \cos\theta) + \dfrac{4l}{27} = \dfrac{40l}{27}$ | M1 A1 | |
**(5 marks) — Total 14 marks**
---6. One end of a light inextensible string of length $l$ is attached to a fixed point $A$. The other end is attached to a particle $P$ of mass $m$ which is hanging freely at rest at point $B$. The particle $P$ is projected horizontally from $B$ with speed $\sqrt { } ( 3 g l )$. When $A P$ makes an angle $\theta$ with the downward vertical and the string remains taut, the tension in the string is $T$.
\begin{enumerate}[label=(\alph*)]
\item Show that $T = m g ( 1 + 3 \cos \theta )$.
\item Find the speed of $P$ at the instant when the string becomes slack.
\item Find the maximum height above the level of $B$ reached by $P$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2006 Q6 [14]}}