| Exam Board | Edexcel |
|---|---|
| Module | M3 (Mechanics 3) |
| Year | 2022 |
| Session | June |
| Marks | 13 |
| 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) requires combining energy and Newton's second law in a non-trivial way; part (b) involves finding when tension becomes zero; part (c) requires projectile motion analysis. The problem demands careful coordination of multiple mechanics principles and algebraic manipulation, placing it above average difficulty but within reach of well-prepared 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 |
|---|---|---|
| Working | Marks | Guidance |
| \(S - mg\cos\theta = \frac{mv^2}{a}\) | M1A1 | Equation of motion along radius. Must have 3 terms with weight resolved. Acceleration in either form. Fully correct equation with \(v^2/r\) |
| \(\frac{1}{2}mv^2 - \frac{1}{2}m \times \frac{9ag}{5} = mga\cos\theta\) | M1A1 | Energy equation from \(A\) to general position. Difference of 2 KE terms and loss of PE. M0 for \(v^2 = u^2 + 2as\). Fully correct equation |
| \(mv^2 = 2mga\cos\theta + \frac{9}{5}mga\) | ||
| \(S = mg\cos\theta + 2mg\cos\theta + \frac{9}{5}mg\) | DM1 | Eliminate \(v^2\) between 2 equations. Depends on both preceding M marks |
| \(S = \frac{3}{5}mg(5\cos\theta + 3)\) | A1* cso (6) | Obtain given result from fully correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| \(S = 0 \Rightarrow \cos\theta = -\frac{3}{5}\) | B1 | \(\cos\theta = -\frac{3}{5}\) seen explicitly or used |
| \(v^2 = \frac{3ag}{5}\), \(v = \sqrt{\frac{3ag}{5}}\) | M1A1* (3) | Use their \(\cos\theta\) to obtain \(v^2\) or \(v\). Correct answer from correct working |
| Answer | Marks | Guidance |
|---|---|---|
| Working | Marks | Guidance |
| Vertical component \(= \sqrt{\frac{3ag}{5}} \times \frac{4}{5}\) | M1 | Use their \(\theta\) and \(v\) to obtain vertical component of velocity (allow sin/cos confusion) |
| Vertical distance to highest point: \(0 = \frac{16}{25} \times \frac{3ag}{5} - 2gs\) | M1 | Correct method to find vertical distance using their vertical component of velocity |
| \(s = \frac{24}{125}a\) | A1 | Correct expression (may be implied) |
| Total distance above \(O = \frac{24}{125}a + \frac{3}{5}a = \frac{99}{125}a \approx 0.79a\) or better | A1ft (4) | Find total distance above \(O\) by adding \(\frac{3a}{5}\) to previous answer. Both M marks needed |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Uses their \(\theta\) and \(v\) to obtain horizontal component at highest point \(\sqrt{\frac{3ag}{5}}\cos\theta\) | M1 | |
| Forms energy equation with 2 KE terms and gain in PE: \(\frac{1}{2}m\frac{3ag}{5} - \frac{1}{2}m\frac{3ag}{5}\left(\frac{3}{5}\right)^2 = mgs\) | M1 | Must have 2 KE terms and gain in PE |
| Correct vertical distance \(s = \frac{24}{125}a\) | A1 | |
| Total distance above \(O\) by adding \(\frac{3a}{5}\): \(\frac{99}{125}a\), or \(0.79a\) or better | A1ft | Both M marks needed |
| Answer | Marks | Guidance |
|---|---|---|
| Answer | Mark | Guidance |
| Uses their \(\theta\) and \(v\) to obtain horizontal component at highest point \(\sqrt{\frac{3ag}{5}}\cos\theta\) | M1 | |
| Forms energy equation with 2 KE terms and gain in PE: \(\frac{1}{2}m\frac{9ag}{5} - \frac{1}{2}m\frac{3ag}{5}\left(\frac{3}{5}\right)^2 = mgh\) | M1, A1 | Must have 2 KE terms and gain in PE |
| Total distance above \(O\): \(h = \frac{99}{125}a\) | A1 | Do not isw |
# Question 6:
## Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| $S - mg\cos\theta = \frac{mv^2}{a}$ | M1A1 | Equation of motion along radius. Must have 3 terms with weight resolved. Acceleration in either form. Fully correct equation with $v^2/r$ |
| $\frac{1}{2}mv^2 - \frac{1}{2}m \times \frac{9ag}{5} = mga\cos\theta$ | M1A1 | Energy equation from $A$ to general position. Difference of 2 KE terms and loss of PE. M0 for $v^2 = u^2 + 2as$. Fully correct equation |
| $mv^2 = 2mga\cos\theta + \frac{9}{5}mga$ | | |
| $S = mg\cos\theta + 2mg\cos\theta + \frac{9}{5}mg$ | DM1 | Eliminate $v^2$ between 2 equations. Depends on both preceding M marks |
| $S = \frac{3}{5}mg(5\cos\theta + 3)$ | A1* cso (6) | Obtain given result from fully correct working |
## Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| $S = 0 \Rightarrow \cos\theta = -\frac{3}{5}$ | B1 | $\cos\theta = -\frac{3}{5}$ seen explicitly or used |
| $v^2 = \frac{3ag}{5}$, $v = \sqrt{\frac{3ag}{5}}$ | M1A1* (3) | Use their $\cos\theta$ to obtain $v^2$ or $v$. Correct answer from correct working |
## Part (c):
| Working | Marks | Guidance |
|---------|-------|----------|
| Vertical component $= \sqrt{\frac{3ag}{5}} \times \frac{4}{5}$ | M1 | Use their $\theta$ and $v$ to obtain vertical component of velocity (allow sin/cos confusion) |
| Vertical distance to highest point: $0 = \frac{16}{25} \times \frac{3ag}{5} - 2gs$ | M1 | Correct method to find vertical distance using their vertical component of velocity |
| $s = \frac{24}{125}a$ | A1 | Correct expression (may be implied) |
| Total distance above $O = \frac{24}{125}a + \frac{3}{5}a = \frac{99}{125}a \approx 0.79a$ or better | A1ft (4) | Find total distance above $O$ by adding $\frac{3a}{5}$ to previous answer. Both M marks needed |
# ALT 1 (Conservation of Energy from slack):
| Answer | Mark | Guidance |
|--------|------|----------|
| Uses their $\theta$ and $v$ to obtain horizontal component at highest point $\sqrt{\frac{3ag}{5}}\cos\theta$ | M1 | |
| Forms energy equation with 2 KE terms and gain in PE: $\frac{1}{2}m\frac{3ag}{5} - \frac{1}{2}m\frac{3ag}{5}\left(\frac{3}{5}\right)^2 = mgs$ | M1 | Must have 2 KE terms and gain in PE |
| Correct vertical distance $s = \frac{24}{125}a$ | A1 | |
| Total distance above $O$ by adding $\frac{3a}{5}$: $\frac{99}{125}a$, or $0.79a$ or better | A1ft | Both M marks needed |
# ALT 2 (Conservation of Energy from initial position A):
| Answer | Mark | Guidance |
|--------|------|----------|
| Uses their $\theta$ and $v$ to obtain horizontal component at highest point $\sqrt{\frac{3ag}{5}}\cos\theta$ | M1 | |
| Forms energy equation with 2 KE terms and gain in PE: $\frac{1}{2}m\frac{9ag}{5} - \frac{1}{2}m\frac{3ag}{5}\left(\frac{3}{5}\right)^2 = mgh$ | M1, A1 | Must have 2 KE terms and gain in PE |
| Total distance above $O$: $h = \frac{99}{125}a$ | A1 | Do not isw |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{2e837bb9-4ada-4f0f-8b21-2730611335f2-20_499_748_244_653}
\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 $a$. The other end of the string is attached to a fixed point $O$. The particle is held at the point $A$, where $O A = a$ and $O A$ is horizontal, as shown in Figure 4.
The particle is projected vertically downwards with speed $\sqrt { \frac { 9 a g } { 5 } }$\\
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 $S$.
\begin{enumerate}[label=(\alph*)]
\item Show that $S = \frac { 3 } { 5 } m g ( 5 \cos \theta + 3 )$
At the instant when the string becomes slack, the speed of $P$ is $v$
\item Show that $v = \sqrt { \frac { 3 a g } { 5 } }$
\item Find the maximum height of $P$ above the horizontal level of $O$
\end{enumerate}
\hfill \mbox{\textit{Edexcel M3 2022 Q6 [13]}}