Edexcel M3 2022 January — Question 7 17 marks

Exam BoardEdexcel
ModuleM3 (Mechanics 3)
Year2022
SessionJanuary
Marks17
PaperDownload PDF ↗
Mark schemeDownload PDF ↗
TopicCircular Motion 2
TypeMaximum/minimum tension or reaction
DifficultyStandard +0.3 This is a standard vertical circular motion problem requiring energy conservation, identification of tension extrema positions, and application of circular motion dynamics. While it involves multiple parts and careful algebraic manipulation, the techniques are routine for M3 students with no novel problem-solving insight required—slightly easier than average A-level.
Spec6.02d Mechanical energy: KE and PE concepts6.02e Calculate KE and PE: using formulae6.05e Radial/tangential acceleration
7. \begin{figure}[h]
\includegraphics[alt={},max width=\textwidth]{bd1e2b07-4a87-49d6-addd-c9f67467ef2f-24_518_538_264_753} \captionsetup{labelformat=empty} \caption{Figure 4}
\end{figure} A particle \(P\) of mass \(m\) is attached to one end of a light rod of length \(l\). The other end of the rod is attached to a fixed point \(O\). The rod can rotate freely in a vertical plane about \(O\). The particle is projected with speed \(u\) from a point \(A\). The line \(O A\) makes an angle \(\alpha\) with the upward vertical through \(O\), where \(\alpha < \frac { \pi } { 2 }\) When \(O P\) makes an angle \(\theta\) with the upward vertical through \(O\), the speed of \(P\) is \(v\), as shown in Figure 4.
  1. Show that \(v ^ { 2 } = u ^ { 2 } - 2 g l ( \cos \theta - \cos \alpha )\) Given that \(\cos \alpha = \frac { 2 } { 5 }\) and that \(u = \sqrt { 3 g l }\)
  2. show that \(P\) moves in a complete vertical circle. As the rod rotates, the least tension in the rod is \(T\) and the greatest tension is \(k T\)
  3. Find the exact value of \(k\)
Question 7:
Part (a):
AnswerMarks Guidance
WorkingMarks Guidance
\(mgl(\cos\theta - \cos\alpha) = \frac{1}{2}mu^2 - \frac{1}{2}mv^2\)M1A1A1 Attempt energy equation from \(A\) to general position; correct gain in PE or loss in KE; fully correct equation
\(v^2 = u^2 - 2gl(\cos\theta - \cos\alpha)\) *A1* Reach given result from fully correct working
Part (b):
AnswerMarks Guidance
WorkingMarks Guidance
\(\cos\alpha = \frac{2}{5}\), \(v^2 = 3gl - 2gl\left(\cos\theta - \frac{2}{5}\right)\)M1A1 Sub \(u=\sqrt{3gl}\) and \(\cos\alpha = \frac{2}{5}\) in result from (a)
At top \(\theta = 0°\): \(v^2 = 3gl - 2gl\times\frac{3}{5}\)M1 Put \(\theta=0\) to find expression for \(v^2\) at top
\(v^2 = \frac{9gl}{5}\)
\(v^2 > 0 \Rightarrow\) complete circle *A1* Fully correct working and conclusion with reason referencing \(v^2\), \(v\), or KE
Part (c):
AnswerMarks Guidance
WorkingMarks Guidance
Equation of motion along radius at lowest point: \(kT - mg = \frac{mw^2}{l}\)M1A1 Form equation of motion at lowest point; correct equation with acceleration in \(v^2/r\) form
\(\theta = 180°\): \(w^2 = 3gl - 2gl\left(-1-\frac{2}{5}\right) = \frac{29gl}{5}\)M1 Use \(\theta=180\) in result from (a) to get expression for \(w^2\)
\(kT = \frac{m}{l}\times\frac{29gl}{5} + mg = \frac{34mg}{5}\)M1A1 Eliminate \(w^2\) and obtain expression for \(kT\); correct expression for \(kT\)
At highest point: \(T_2 + mg = \frac{mv^2}{l}\)M1 Form equation of motion at highest point
\(\theta = 0\): \(T_2 = \frac{9mg}{5} - mg = \frac{4mg}{5}\)M1A1 Sub \(\theta=0\); correct expression for \(T\)
\(k\cdot\frac{4mg}{5} = \frac{34mg}{5} \Rightarrow k = \frac{17}{2}\)A1 Correct value of \(k\); must be exact
Question 7(c) ALT1:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
Better equation of motion at top or bottom: \(T - mg = \frac{mv^2}{l}\) or \(T + mg = \frac{mv^2}{l}\)M1 A1 Either correct equation of motion at top or bottom
Other equation of motion – see aboveM1 Second equation of motion
Finding speed at the bottom: \(\theta = 180\), \(w^2 = 3gl - 2gl\left(-1 - \frac{2}{5}\right) = \frac{29gl}{5}\)M1 Use of energy/velocity equation
Finding maximum tension (lowest point): \(\theta = 180\), \(T = \frac{m}{l} \times \frac{29gl}{5} + mg = \frac{34mg}{5}\)M1 A1
Finding minimum tension (highest point): \(\theta = 0\), \(T = \frac{9mg}{5} - mg = \frac{4mg}{5}\)M1 A1
Dividing tensions to reach correct answer: \(k\frac{4mg}{5} = \frac{34mg}{5} \Rightarrow k = \frac{17}{2}\)A1
Question 7(c) ALT2:
AnswerMarks Guidance
Answer/WorkingMarks Guidance
General equation of motion: \(T + mg\cos\theta = \frac{mv^2}{l}\)M1 A1
Use of \(u = \sqrt{3gl}\) and \(\cos\alpha = \frac{2}{5}\) to replace \(v^2\) in equation of motionM1
Finding speed at lowest point: \(\theta = 180\), \(w^2 = 3gl - 2gl\left(-1-\frac{2}{5}\right) = \frac{29gl}{5}\)M1
Finding maximum tension (lowest point): \(\theta = 180\), \(T = \frac{m}{l} \times \frac{29gl}{5} + mg = \frac{34mg}{5}\)M1 A1
Finding minimum tension (highest point): \(\theta = 0\), \(T = \frac{9mg}{5} - mg = \frac{4mg}{5}\)M1 A1
Dividing tensions to reach correct answer: \(k\frac{4mg}{5} = \frac{34mg}{5} \Rightarrow k = \frac{17}{2}\)A1 
# Question 7:

## Part (a):
| Working | Marks | Guidance |
|---------|-------|----------|
| $mgl(\cos\theta - \cos\alpha) = \frac{1}{2}mu^2 - \frac{1}{2}mv^2$ | M1A1A1 | Attempt energy equation from $A$ to general position; correct gain in PE or loss in KE; fully correct equation |
| $v^2 = u^2 - 2gl(\cos\theta - \cos\alpha)$ * | A1* | Reach **given** result from fully correct working |

## Part (b):
| Working | Marks | Guidance |
|---------|-------|----------|
| $\cos\alpha = \frac{2}{5}$, $v^2 = 3gl - 2gl\left(\cos\theta - \frac{2}{5}\right)$ | M1A1 | Sub $u=\sqrt{3gl}$ and $\cos\alpha = \frac{2}{5}$ in result from (a) |
| At top $\theta = 0°$: $v^2 = 3gl - 2gl\times\frac{3}{5}$ | M1 | Put $\theta=0$ to find expression for $v^2$ at top |
| $v^2 = \frac{9gl}{5}$ | | |
| $v^2 > 0 \Rightarrow$ complete circle * | A1* | Fully correct working and conclusion with reason referencing $v^2$, $v$, or KE |

## Part (c):
| Working | Marks | Guidance |
|---------|-------|----------|
| Equation of motion along radius at lowest point: $kT - mg = \frac{mw^2}{l}$ | M1A1 | Form equation of motion at lowest point; correct equation with acceleration in $v^2/r$ form |
| $\theta = 180°$: $w^2 = 3gl - 2gl\left(-1-\frac{2}{5}\right) = \frac{29gl}{5}$ | M1 | Use $\theta=180$ in result from (a) to get expression for $w^2$ |
| $kT = \frac{m}{l}\times\frac{29gl}{5} + mg = \frac{34mg}{5}$ | M1A1 | Eliminate $w^2$ and obtain expression for $kT$; correct expression for $kT$ |
| At highest point: $T_2 + mg = \frac{mv^2}{l}$ | M1 | Form equation of motion at highest point |
| $\theta = 0$: $T_2 = \frac{9mg}{5} - mg = \frac{4mg}{5}$ | M1A1 | Sub $\theta=0$; correct expression for $T$ |
| $k\cdot\frac{4mg}{5} = \frac{34mg}{5} \Rightarrow k = \frac{17}{2}$ | A1 | Correct value of $k$; must be exact |

## Question 7(c) ALT1:

| Answer/Working | Marks | Guidance |
|---|---|---|
| Better equation of motion at top **or** bottom: $T - mg = \frac{mv^2}{l}$ **or** $T + mg = \frac{mv^2}{l}$ | M1 A1 | Either correct equation of motion at top or bottom |
| Other equation of motion – see above | M1 | Second equation of motion |
| Finding speed at the bottom: $\theta = 180$, $w^2 = 3gl - 2gl\left(-1 - \frac{2}{5}\right) = \frac{29gl}{5}$ | M1 | Use of energy/velocity equation |
| Finding maximum tension (lowest point): $\theta = 180$, $T = \frac{m}{l} \times \frac{29gl}{5} + mg = \frac{34mg}{5}$ | M1 A1 | |
| Finding minimum tension (highest point): $\theta = 0$, $T = \frac{9mg}{5} - mg = \frac{4mg}{5}$ | M1 A1 | |
| Dividing tensions to reach correct answer: $k\frac{4mg}{5} = \frac{34mg}{5} \Rightarrow k = \frac{17}{2}$ | A1 | |

---

## Question 7(c) ALT2:

| Answer/Working | Marks | Guidance |
|---|---|---|
| General equation of motion: $T + mg\cos\theta = \frac{mv^2}{l}$ | M1 A1 | |
| Use of $u = \sqrt{3gl}$ and $\cos\alpha = \frac{2}{5}$ to replace $v^2$ in equation of motion | M1 | |
| Finding speed at lowest point: $\theta = 180$, $w^2 = 3gl - 2gl\left(-1-\frac{2}{5}\right) = \frac{29gl}{5}$ | M1 | |
| Finding maximum tension (lowest point): $\theta = 180$, $T = \frac{m}{l} \times \frac{29gl}{5} + mg = \frac{34mg}{5}$ | M1 A1 | |
| Finding minimum tension (highest point): $\theta = 0$, $T = \frac{9mg}{5} - mg = \frac{4mg}{5}$ | M1 A1 | |
| Dividing tensions to reach correct answer: $k\frac{4mg}{5} = \frac{34mg}{5} \Rightarrow k = \frac{17}{2}$ | A1 | |
7.

\begin{figure}[h]
\begin{center}
  \includegraphics[alt={},max width=\textwidth]{bd1e2b07-4a87-49d6-addd-c9f67467ef2f-24_518_538_264_753}
\captionsetup{labelformat=empty}
\caption{Figure 4}
\end{center}
\end{figure}

A particle $P$ of mass $m$ is attached to one end of a light rod of length $l$. The other end of the rod is attached to a fixed point $O$. The rod can rotate freely in a vertical plane about $O$. The particle is projected with speed $u$ from a point $A$. The line $O A$ makes an angle $\alpha$ with the upward vertical through $O$, where $\alpha < \frac { \pi } { 2 }$

When $O P$ makes an angle $\theta$ with the upward vertical through $O$, the speed of $P$ is $v$, as shown in Figure 4.
\begin{enumerate}[label=(\alph*)]
\item Show that $v ^ { 2 } = u ^ { 2 } - 2 g l ( \cos \theta - \cos \alpha )$

Given that $\cos \alpha = \frac { 2 } { 5 }$ and that $u = \sqrt { 3 g l }$
\item show that $P$ moves in a complete vertical circle.

As the rod rotates, the least tension in the rod is $T$ and the greatest tension is $k T$
\item Find the exact value of $k$
\end{enumerate}

\hfill \mbox{\textit{Edexcel M3 2022 Q7 [17]}}
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Q1 7 Q2 8 Q3 8 Q4 8 Q5 12 Q6 15 Q7 17