| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2023 |
| Session | June |
| Marks | 16 |
| 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 requiring energy conservation and circular motion equations (T = mv²/r ± mg component). Part (a) is a structured 'show that' guiding students through the method. Parts (b) and (c) require recognizing that slack occurs when T=0, then applying projectile motion—all well-practiced techniques in M3 with no novel insight needed. Slightly above average due to the multi-step nature and algebraic manipulation, but follows a predictable template. |
| 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 |
| \(\frac{1}{2}mv^2 - \frac{1}{2}mu^2 = mga\sin\theta\) | M1 A1 A1 | M1: energy equation with correct no. of terms, dim correct. May use \(h\) instead of \(a\sin\theta\). A1: correct equation with at most one error |
| \(T - mg\sin\theta = \frac{mv^2}{a}\) | M1 A1 A1 | M1: equation of motion towards \(O\), correct no. of terms, condone sign errors and sin/cos confusion. Accept acceleration in either circular form but not '\(a\)'. Radius may be given as \(r\) |
| \(T = \frac{mu^2}{a} + 3mg\sin\theta\) | A1* | Given answer correctly obtained and written exactly as printed |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(0 = \frac{m\left(\frac{12ag}{5}\right)}{a} + 3mg\sin\theta\) | M1 | Put \(T = 0\) and \(u = 2\sqrt{\frac{3ag}{5}}\) |
| \(\sin\theta = -\frac{4}{5}\) | A1 | Correct value of \(\sin\theta\) |
| \(\frac{1}{2}mv^2 - \frac{1}{2}m\left(\frac{12ag}{5}\right) = mga \times -\frac{4}{5}\) OR \(0 - mg \times -\frac{4}{5} = \frac{mv^2}{a}\) | M1 | Put \(u = 2\sqrt{\frac{3ag}{5}}\) and their \(\sin\theta\) into energy equation, OR put \(T=0\) and their \(\sin\theta\) into equation of motion |
| \(v = 2\sqrt{\frac{ag}{5}}\), \(0.89\sqrt{ag}\) or better | A1 | Correct answer |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| Vertical motion: \(0 = \left(2\sqrt{\frac{ag}{5}} \times \frac{3}{5}\right)^2 - 2gh\) OR Energy: \(mgh = \frac{1}{2}m\left(2\sqrt{\frac{ag}{5}}\right)^2 - \frac{1}{2}m\left(2\sqrt{\frac{ag}{5}} \times \frac{4}{5}\right)^2\) | M1 A1ft | M1: use vertical motion or energy to obtain equation in \(h\) only; a component of speed must be used. A1ft: correct equation ft on their answer to (b) |
| \(h = \frac{18a}{125}\) | A1 | Correct value of \(h\) |
| \(H = h - a\sin\theta = \frac{18a}{125} + \frac{4a}{5}\) | dM1 | Correct method to find \(H\); dependent on previous M |
| \(H = \frac{118a}{125}\), \(0.94a\), \(0.944a\) | A1 | cao |
| Answer | Marks | Guidance |
|---|---|---|
| Answer/Working | Mark | Guidance |
| \(mgH = \frac{1}{2}m\left\{\frac{12ag}{5} - \left(2\sqrt{\frac{ag}{5}} \times \frac{4}{5}\right)^2\right\}\) | M2 A1ft A1 | M2: complete method to obtain equation in \(H\) only, must use horizontal component of velocity at top |
| \(H = \frac{118a}{125}\), \(0.94a\), \(0.944a\) | A1 | Correct answer |
## Question 6(a):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $\frac{1}{2}mv^2 - \frac{1}{2}mu^2 = mga\sin\theta$ | M1 A1 A1 | M1: energy equation with correct no. of terms, dim correct. May use $h$ instead of $a\sin\theta$. A1: correct equation with at most one error |
| $T - mg\sin\theta = \frac{mv^2}{a}$ | M1 A1 A1 | M1: equation of motion towards $O$, correct no. of terms, condone sign errors and sin/cos confusion. Accept acceleration in either circular form but not '$a$'. Radius may be given as $r$ |
| $T = \frac{mu^2}{a} + 3mg\sin\theta$ | A1* | Given answer correctly obtained and written exactly as printed |
---
## Question 6(b):
| Answer/Working | Mark | Guidance |
|---|---|---|
| $0 = \frac{m\left(\frac{12ag}{5}\right)}{a} + 3mg\sin\theta$ | M1 | Put $T = 0$ and $u = 2\sqrt{\frac{3ag}{5}}$ |
| $\sin\theta = -\frac{4}{5}$ | A1 | Correct value of $\sin\theta$ |
| $\frac{1}{2}mv^2 - \frac{1}{2}m\left(\frac{12ag}{5}\right) = mga \times -\frac{4}{5}$ OR $0 - mg \times -\frac{4}{5} = \frac{mv^2}{a}$ | M1 | Put $u = 2\sqrt{\frac{3ag}{5}}$ and their $\sin\theta$ into energy equation, OR put $T=0$ and their $\sin\theta$ into equation of motion |
| $v = 2\sqrt{\frac{ag}{5}}$, $0.89\sqrt{ag}$ or better | A1 | Correct answer |
---
## Question 6(c):
| Answer/Working | Mark | Guidance |
|---|---|---|
| Vertical motion: $0 = \left(2\sqrt{\frac{ag}{5}} \times \frac{3}{5}\right)^2 - 2gh$ OR Energy: $mgh = \frac{1}{2}m\left(2\sqrt{\frac{ag}{5}}\right)^2 - \frac{1}{2}m\left(2\sqrt{\frac{ag}{5}} \times \frac{4}{5}\right)^2$ | M1 A1ft | M1: use vertical motion or energy to obtain equation in $h$ only; a component of speed must be used. A1ft: correct equation ft on their answer to (b) |
| $h = \frac{18a}{125}$ | A1 | Correct value of $h$ |
| $H = h - a\sin\theta = \frac{18a}{125} + \frac{4a}{5}$ | dM1 | Correct method to find $H$; dependent on previous M |
| $H = \frac{118a}{125}$, $0.94a$, $0.944a$ | A1 | cao |
**OR using energy from start to top:**
| Answer/Working | Mark | Guidance |
|---|---|---|
| $mgH = \frac{1}{2}m\left\{\frac{12ag}{5} - \left(2\sqrt{\frac{ag}{5}} \times \frac{4}{5}\right)^2\right\}$ | M2 A1ft A1 | M2: complete method to obtain equation in $H$ only, must use horizontal component of velocity at top |
| $H = \frac{118a}{125}$, $0.94a$, $0.944a$ | A1 | Correct answer |6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{dceb2432-117c-40fe-bf3d-782beeb42e41-20_134_653_243_707}
\captionsetup{labelformat=empty}
\caption{Figure 5}
\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 $O$. The particle $P$ is held at rest with the string taut and horizontal and is then projected vertically downwards with speed $u$, as shown in Figure 5.
Air resistance is modelled as being negligible.\\
At the instant when the string has turned through an angle $\theta$ and the string is taut, the tension in the string is $T$.
\begin{enumerate}[label=(\alph*)]
\item Show that $T = \frac { m u ^ { 2 } } { a } + 3 m g \sin \theta$
Given that $u = 2 \sqrt { \frac { 3 a g } { 5 } }$
\item find, in terms of $a$ and $g$, the speed of $P$ at the instant when the string goes slack.
\item Hence find, in terms of $a$, the maximum height of $P$ above $O$ in the subsequent motion.
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2023 Q6 [16]}}