| Exam Board | Edexcel |
|---|---|
| Module | FM2 (Further Mechanics 2) |
| Year | 2020 |
| 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 standard Further Mechanics 2 vertical circle problem requiring energy conservation, circular motion dynamics (T = mv²/r ± mg component), and projectile motion after the string slackens. Part (a) is a guided 'show that' requiring two equations, parts (b-c) involve finding when T=0 and subsequent projectile motion. While multi-step, these are well-practiced FM2 techniques with no novel insight required, placing it moderately above average difficulty. |
| Spec | 3.02h Motion under gravity: vector form6.02i Conservation of energy: mechanical energy principle6.05e Radial/tangential acceleration |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Conservation of energy | M1 | 3.1a - Must include all terms; condone sign errors and sin/cos confusion |
| \(\frac{1}{2}mu^2 + mgl\sin\alpha = \frac{1}{2}mv^2\), \(\left(v^2 = \frac{9gl}{5} + 2gl\sin\alpha\right)\) | A1 | 1.1b - Correct unsimplified equation |
| Equation of motion | M1 | 3.1a - Must include all terms; condone sign errors and sin/cos confusion |
| \(T - mg\sin\alpha = \frac{mv^2}{l}\) | A1 | 1.1b - Correct unsimplified equation |
| Complete strategy to find \(T\) in terms of \(\alpha\) | M1 | 2.1 - Complete strategy using conservation of energy and circular motion to form sufficient equations |
| \(T = mg\sin\alpha + \frac{mv^2}{l} = mg\sin\alpha + \frac{9mg}{5} + 2mg\sin\alpha = 3mg\sin\alpha + \frac{9mg}{5}\) * | A1* | 2.2a - Obtain given answer from correct working |
| Total: (6) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| String slack \(\Rightarrow T = 0 \Rightarrow \sin\alpha = -\frac{3}{5}\) | B1 | 3.1a - Correct deduction |
| Use energy equation to find \(v\): \(v^2 = \frac{9gl}{5} - \frac{3}{5} \times 2gl\), \(v = \sqrt{\frac{3gl}{5}}\) | M1, A1 | 1.1b - Substitute value to find \(v^2\); correct only |
| Total: (3) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Initial vertical component of speed \(= \frac{4}{5}\times\sqrt{\frac{3gl}{5}}\) | B1 | 1.1b - Correct vertical component of velocity when string goes slack |
| Use of suvat: \(0 = u^2 - 2gh = \frac{16}{25} \times \frac{3gl}{5} - 2gh\) | M1 | 3.1a - Use of \(v^2 = u^2 + 2as\) or alternative complete method |
| \(h = \frac{24l}{125}\) | A1 | 1.1b - Additional height correct |
| Total height above \(O = \frac{3l}{5} + \frac{24l}{125} = \frac{99l}{125}\) | A1 | 2.2a - Total height correct |
| Total: (4) |
| Answer | Marks | Guidance |
|---|---|---|
| Working/Answer | Mark | Guidance |
| Initial horizontal component of speed \(= \frac{3}{5}\times\sqrt{\frac{3gl}{5}}\) | B1 | 1.1b - Correct horizontal component of velocity when string goes slack |
| Conservation of energy: \(mgh = \frac{1}{2}m\left(\frac{9}{5}\right)gl - \frac{1}{2}m\left(\frac{9}{25}\times\frac{3gl}{5}\right)\) | M1, A1 | 3.1a, 1.1b - Use of conservation of energy; all terms required; condone sign errors; correct unsimplified equation in \(h\) and \(l\) |
| \(h = \frac{99l}{125}\) | A1 | 2.2a - Correct answer |
| Total: (4) |
# Question 6(a):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Conservation of energy | M1 | 3.1a - Must include all terms; condone sign errors and sin/cos confusion |
| $\frac{1}{2}mu^2 + mgl\sin\alpha = \frac{1}{2}mv^2$, $\left(v^2 = \frac{9gl}{5} + 2gl\sin\alpha\right)$ | A1 | 1.1b - Correct unsimplified equation |
| Equation of motion | M1 | 3.1a - Must include all terms; condone sign errors and sin/cos confusion |
| $T - mg\sin\alpha = \frac{mv^2}{l}$ | A1 | 1.1b - Correct unsimplified equation |
| Complete strategy to find $T$ in terms of $\alpha$ | M1 | 2.1 - Complete strategy using conservation of energy and circular motion to form sufficient equations |
| $T = mg\sin\alpha + \frac{mv^2}{l} = mg\sin\alpha + \frac{9mg}{5} + 2mg\sin\alpha = 3mg\sin\alpha + \frac{9mg}{5}$ * | A1* | 2.2a - Obtain given answer from correct working |
| **Total: (6)** | | |
---
# Question 6(b):
| Working/Answer | Mark | Guidance |
|---|---|---|
| String slack $\Rightarrow T = 0 \Rightarrow \sin\alpha = -\frac{3}{5}$ | B1 | 3.1a - Correct deduction |
| Use energy equation to find $v$: $v^2 = \frac{9gl}{5} - \frac{3}{5} \times 2gl$, $v = \sqrt{\frac{3gl}{5}}$ | M1, A1 | 1.1b - Substitute value to find $v^2$; correct only |
| **Total: (3)** | | |
---
# Question 6(c):
| Working/Answer | Mark | Guidance |
|---|---|---|
| Initial vertical component of speed $= \frac{4}{5}\times\sqrt{\frac{3gl}{5}}$ | B1 | 1.1b - Correct vertical component of velocity when string goes slack |
| Use of suvat: $0 = u^2 - 2gh = \frac{16}{25} \times \frac{3gl}{5} - 2gh$ | M1 | 3.1a - Use of $v^2 = u^2 + 2as$ or alternative complete method |
| $h = \frac{24l}{125}$ | A1 | 1.1b - Additional height correct |
| Total height above $O = \frac{3l}{5} + \frac{24l}{125} = \frac{99l}{125}$ | A1 | 2.2a - Total height correct |
| **Total: (4)** | | |
---
# Question 6(c) alt:
| Working/Answer | Mark | Guidance |
|---|---|---|
| Initial horizontal component of speed $= \frac{3}{5}\times\sqrt{\frac{3gl}{5}}$ | B1 | 1.1b - Correct horizontal component of velocity when string goes slack |
| Conservation of energy: $mgh = \frac{1}{2}m\left(\frac{9}{5}\right)gl - \frac{1}{2}m\left(\frac{9}{25}\times\frac{3gl}{5}\right)$ | M1, A1 | 3.1a, 1.1b - Use of conservation of energy; all terms required; condone sign errors; correct unsimplified equation in $h$ and $l$ |
| $h = \frac{99l}{125}$ | A1 | 2.2a - Correct answer |
| **Total: (4)** | | |
---6.
\begin{figure}[h]
\begin{center}
\includegraphics[alt={},max width=\textwidth]{962c2b40-3c45-4eed-a0af-a59068bda0e1-20_533_543_242_760}
\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 $l$. The other end of the string is attached to a fixed point $O$. The particle is held with the string taut and $O P$ horizontal. The particle is then projected vertically downwards with speed $u$, where $u ^ { 2 } = \frac { 9 } { 5 } \mathrm { gl }$. When $O P$ has turned through an angle $\alpha$ and the string is still taut, the speed of $P$ is $v$, as shown in Figure 5. At this instant the tension in the string is $T$.
\begin{enumerate}[label=(\alph*)]
\item Show that $T = 3 m g \sin \alpha + \frac { 9 } { 5 } m g$
\item Find, in terms of $g$ and $l$, the speed of $P$ at the instant when the string goes slack.
\item Find, in terms of $l$, the greatest vertical height reached by $P$ above the level of $O$.
\end{enumerate}
\hfill \mbox{\textit{Edexcel FM2 2020 Q6 [13]}}